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Math Problems
Algebra 1
Solve a quadratic equation by factoring
Solve for
x
x
x
. Enter the solutions from least to greatest.
\newline
3
x
2
+
36
x
+
81
=
0
lesser
x
=
□
greater
x
=
□
\begin{array}{l} 3 x^{2}+36 x+81=0 \\ \text { lesser } x=\square \\ \text { greater } x=\square \end{array}
3
x
2
+
36
x
+
81
=
0
lesser
x
=
□
greater
x
=
□
Get tutor help
Solve for
x
x
x
. Enter the solutions from least to greatest.
\newline
2
x
2
−
24
x
+
54
=
0
2 x^{2}-24 x+54=0
2
x
2
−
24
x
+
54
=
0
\newline
lesser
x
=
x=
x
=
\newline
greater
x
=
x=
x
=
Get tutor help
Solve for
x
x
x
. Enter the solutions from least to greatest.
\newline
4
x
2
+
40
x
+
84
=
0
lesser
x
=
□
greater
x
=
□
\begin{array}{l} 4 x^{2}+40 x+84=0 \\ \text { lesser } x=\square \\ \text { greater } x=\square \end{array}
4
x
2
+
40
x
+
84
=
0
lesser
x
=
□
greater
x
=
□
Get tutor help
Solve for
x
x
x
. Enter the solutions from least to greatest.
\newline
x
2
−
x
−
12
=
0
lesser
x
=
□
greater
x
=
□
\begin{array}{l} x^{2}-x-12=0 \\ \text { lesser } x=\square \\ \text { greater } x=\square \end{array}
x
2
−
x
−
12
=
0
lesser
x
=
□
greater
x
=
□
Get tutor help
Find one value of
x
x
x
that is a solution to the equation:
\newline
(
3
x
+
5
)
2
+
5
(
3
x
+
5
)
+
6
=
0
x
=
□
\begin{array}{l} (3 x+5)^{2}+5(3 x+5)+6=0 \\ x=\square \end{array}
(
3
x
+
5
)
2
+
5
(
3
x
+
5
)
+
6
=
0
x
=
□
Get tutor help
Solve for
x
x
x
. Enter the solutions from least to greatest.
\newline
2
x
2
−
16
x
+
14
=
0
lesser
x
=
□
greater
x
=
□
\begin{array}{l} 2 x^{2}-16 x+14=0 \\ \text { lesser } x=\square \\ \text { greater } x=\square \end{array}
2
x
2
−
16
x
+
14
=
0
lesser
x
=
□
greater
x
=
□
Get tutor help
Solve for
x
x
x
.
\newline
x
2
−
12
x
+
36
=
0
x
=
□
\begin{array}{l} x^{2}-12 x+36=0 \\ x=\square \end{array}
x
2
−
12
x
+
36
=
0
x
=
□
Get tutor help
Solve for
x
x
x
. Enter the solutions from least to greatest.
\newline
4
x
2
+
4
x
−
168
=
0
lesser
x
=
□
greater
x
=
□
\begin{array}{l} 4 x^{2}+4 x-168=0 \\ \text { lesser } x=\square \\ \text { greater } x=\square \end{array}
4
x
2
+
4
x
−
168
=
0
lesser
x
=
□
greater
x
=
□
Get tutor help
Solve for
x
x
x
.
\newline
3
x
2
+
18
x
+
27
=
0
x
=
□
\begin{array}{l} 3 x^{2}+18 x+27=0 \\ x=\square \end{array}
3
x
2
+
18
x
+
27
=
0
x
=
□
Get tutor help
Find one value of
x
x
x
that is a solution to the equation:
\newline
(
5
x
+
2
)
2
+
15
x
+
6
=
0
x
=
□
\begin{array}{l} (5 x+2)^{2}+15 x+6=0 \\ x=\square \end{array}
(
5
x
+
2
)
2
+
15
x
+
6
=
0
x
=
□
Get tutor help
Solve for
x
x
x
.
\newline
x
2
−
2
x
+
1
=
0
x^{2}-2 x+1=0
x
2
−
2
x
+
1
=
0
\newline
x
=
x=
x
=
Get tutor help
Solve for
x
x
x
.
\newline
5
x
2
+
60
x
+
180
=
0
x
=
□
\begin{array}{l} 5 x^{2}+60 x+180=0 \\ x=\square \end{array}
5
x
2
+
60
x
+
180
=
0
x
=
□
Get tutor help
Solve for
x
x
x
. Enter the solutions from least to greatest.
\newline
x
2
−
3
x
−
40
=
0
lesser
x
=
□
greater
x
=
□
\begin{array}{l} x^{2}-3 x-40=0 \\ \text { lesser } x=\square \\ \text { greater } x=\square \end{array}
x
2
−
3
x
−
40
=
0
lesser
x
=
□
greater
x
=
□
Get tutor help
Solve for
x
x
x
. Enter the solutions from least to greatest.
\newline
x
2
−
4
x
+
3
=
0
lesser
x
=
□
greater
x
=
□
\begin{array}{l} x^{2}-4 x+3=0 \\ \text { lesser } x=\square \\ \text { greater } x=\square \end{array}
x
2
−
4
x
+
3
=
0
lesser
x
=
□
greater
x
=
□
Get tutor help
Let
m
=
5
x
−
2
m=5x-2
m
=
5
x
−
2
. Which equation is equivalent to
(
5
x
−
2
)
2
+
35
x
−
14
=
−
12
(5x-2)^{2}+35x-14=-12
(
5
x
−
2
)
2
+
35
x
−
14
=
−
12
in terms of
m
m
m
? Choose
1
1
1
answer:
\newline
(A)
m
2
−
7
m
−
2
=
0
m^{2}-7m-2=0
m
2
−
7
m
−
2
=
0
\newline
(B)
m
2
−
7
m
+
12
=
0
m^{2}-7m+12=0
m
2
−
7
m
+
12
=
0
\newline
(C)
m
2
+
7
m
−
2
=
0
m^{2}+7m-2=0
m
2
+
7
m
−
2
=
0
\newline
(D)
m
2
+
7
m
+
12
=
0
m^{2}+7m+12=0
m
2
+
7
m
+
12
=
0
Get tutor help
Let
m
=
6
x
+
5
m=6x+5
m
=
6
x
+
5
. Which equation is equivalent to
(
6
x
+
5
)
2
−
10
=
−
18
x
−
15
(6x+5)^{2}-10=-18x-15
(
6
x
+
5
)
2
−
10
=
−
18
x
−
15
in terms of
m
m
m
? Choose
1
1
1
answer: (A)
m
2
−
3
m
+
5
=
0
m^{2}-3m+5=0
m
2
−
3
m
+
5
=
0
(B)
m
2
+
3
m
+
5
=
0
m^{2}+3m+5=0
m
2
+
3
m
+
5
=
0
(C)
m
2
−
3
m
−
10
=
0
m^{2}-3m-10=0
m
2
−
3
m
−
10
=
0
(D)
m
2
+
3
m
−
10
=
0
m^{2}+3m-10=0
m
2
+
3
m
−
10
=
0
Get tutor help
Let
p
=
x
2
−
7
p=x^{2}-7
p
=
x
2
−
7
. Which equation is equivalent to
(
x
2
−
7
)
2
−
4
x
2
+
28
=
5
(x^{2}-7)^{2}-4x^{2}+28=5
(
x
2
−
7
)
2
−
4
x
2
+
28
=
5
in terms of
p
p
p
? Choose
1
1
1
answer:
\newline
(A)
p
2
−
4
p
+
23
=
0
p^{2}-4p+23=0
p
2
−
4
p
+
23
=
0
\newline
(B)
p
2
+
4
p
−
5
=
0
p^{2}+4p-5=0
p
2
+
4
p
−
5
=
0
\newline
(C)
p
2
−
4
p
−
5
=
0
p^{2}-4p-5=0
p
2
−
4
p
−
5
=
0
\newline
(D)
p
2
+
4
p
+
23
=
0
p^{2}+4p+23=0
p
2
+
4
p
+
23
=
0
Get tutor help
Let
m
=
2
x
+
3
m=2x+3
m
=
2
x
+
3
. Which equation is equivalent to
(
2
x
+
3
)
2
−
14
x
−
21
=
−
6
(2x+3)^{2}-14x-21=-6
(
2
x
+
3
)
2
−
14
x
−
21
=
−
6
in terms of
m
m
m
? Choose
1
1
1
answer:
\newline
(A)
m
2
+
7
m
+
6
=
0
m^{2}+7m+6=0
m
2
+
7
m
+
6
=
0
\newline
(B)
m
2
−
7
m
−
15
=
0
m^{2}-7m-15=0
m
2
−
7
m
−
15
=
0
\newline
(C)
m
2
−
7
m
+
6
=
0
m^{2}-7m+6=0
m
2
−
7
m
+
6
=
0
\newline
(D)
m
2
+
7
m
−
15
=
0
m^{2}+7m-15=0
m
2
+
7
m
−
15
=
0
Get tutor help
Let
m
=
x
2
+
3
m=x^{2}+3
m
=
x
2
+
3
. Which equation is equivalent to
(
x
2
+
3
)
2
+
7
x
2
+
21
=
−
10
(x^{2}+3)^{2}+7x^{2}+21=-10
(
x
2
+
3
)
2
+
7
x
2
+
21
=
−
10
in terms of
m
m
m
? Choose
1
1
1
answer:
\newline
(A)
m
2
−
7
m
+
31
=
0
m^{2}-7m+31=0
m
2
−
7
m
+
31
=
0
\newline
(B)
m
2
+
7
m
+
31
=
0
m^{2}+7m+31=0
m
2
+
7
m
+
31
=
0
\newline
(C)
m
2
−
7
m
+
10
=
0
m^{2}-7m+10=0
m
2
−
7
m
+
10
=
0
\newline
(D)
m
2
+
7
m
+
10
=
0
m^{2}+7m+10=0
m
2
+
7
m
+
10
=
0
Get tutor help
Let
p
=
x
2
+
6
p = x^{2} + 6
p
=
x
2
+
6
. Which equation is equivalent to
(
x
2
+
6
)
2
−
21
=
4
x
2
+
24
(x^{2} + 6)^{2} - 21 = 4x^{2} + 24
(
x
2
+
6
)
2
−
21
=
4
x
2
+
24
in terms of
p
p
p
? Choose
1
1
1
answer:
\newline
(A)
p
2
+
4
p
−
21
=
0
p^{2} + 4p - 21 = 0
p
2
+
4
p
−
21
=
0
\newline
(B)
p
2
−
4
p
−
21
=
0
p^{2} - 4p - 21 = 0
p
2
−
4
p
−
21
=
0
\newline
(C)
p
2
+
4
p
−
45
=
0
p^{2} + 4p - 45 = 0
p
2
+
4
p
−
45
=
0
\newline
(D)
p
2
−
4
p
−
45
=
0
p^{2} - 4p - 45 = 0
p
2
−
4
p
−
45
=
0
Get tutor help
Let
p
=
4
x
−
7
p=4x-7
p
=
4
x
−
7
. Which equation is equivalent to
(
4
x
−
7
)
2
+
16
=
40
x
−
70
(4x-7)^{2}+16=40x-70
(
4
x
−
7
)
2
+
16
=
40
x
−
70
in terms of
p
p
p
? Choose
1
1
1
answer:
\newline
(A)
p
2
−
10
p
+
86
=
0
p^{2}-10p+86=0
p
2
−
10
p
+
86
=
0
\newline
(B)
p
2
−
10
p
+
16
=
0
p^{2}-10p+16=0
p
2
−
10
p
+
16
=
0
\newline
(C)
p
2
+
10
p
+
86
=
0
p^{2}+10p+86=0
p
2
+
10
p
+
86
=
0
\newline
(D)
p
2
+
10
p
+
16
=
0
p^{2}+10p+16=0
p
2
+
10
p
+
16
=
0
Get tutor help
Find one value of
x
x
x
that is a solution to the equation:
\newline
(
2
x
+
3
)
2
−
4
(
2
x
+
3
)
−
12
=
0
(2x+3)^{2}-4(2x+3)-12=0
(
2
x
+
3
)
2
−
4
(
2
x
+
3
)
−
12
=
0
\newline
x
=
x=
x
=
Get tutor help
Let
m
=
x
2
−
5
m=x^{2}-5
m
=
x
2
−
5
. Which equation is equivalent to
(
x
2
−
5
)
2
−
3
x
2
+
15
=
−
2
(x^{2}-5)^{2}-3x^{2}+15=-2
(
x
2
−
5
)
2
−
3
x
2
+
15
=
−
2
in terms of
m
m
m
? Choose
1
1
1
answer:
\newline
(A)
m
2
+
3
m
+
2
=
0
m^{2}+3m+2=0
m
2
+
3
m
+
2
=
0
\newline
(B)
m
2
+
3
m
+
17
=
0
m^{2}+3m+17=0
m
2
+
3
m
+
17
=
0
\newline
(C)
m
2
−
3
m
+
2
=
0
m^{2}-3m+2=0
m
2
−
3
m
+
2
=
0
\newline
(D)
m
2
−
3
m
+
17
=
0
m^{2}-3m+17=0
m
2
−
3
m
+
17
=
0
Get tutor help
Let
p
=
x
2
−
2
p=x^{2}-2
p
=
x
2
−
2
. Which equation is equivalent to
(
x
2
−
2
)
2
+
18
=
9
x
2
−
18
(x^{2}-2)^{2}+18=9x^{2}-18
(
x
2
−
2
)
2
+
18
=
9
x
2
−
18
in terms of
p
p
p
? Choose
1
1
1
answer:
\newline
(A)
p
2
+
9
p
+
36
=
0
p^{2}+9p+36=0
p
2
+
9
p
+
36
=
0
\newline
(B)
p
2
−
9
p
+
36
=
0
p^{2}-9p+36=0
p
2
−
9
p
+
36
=
0
\newline
(C)
p
2
−
9
p
+
18
=
0
p^{2}-9p+18=0
p
2
−
9
p
+
18
=
0
\newline
(D)
p
2
+
9
p
+
18
=
0
p^{2}+9p+18=0
p
2
+
9
p
+
18
=
0
Get tutor help
Let
p
=
3
x
+
4
p=3x+4
p
=
3
x
+
4
. Which equation is equivalent to
(
3
x
+
4
)
2
−
36
=
15
x
+
20
(3x+4)^{2}-36=15x+20
(
3
x
+
4
)
2
−
36
=
15
x
+
20
in terms of
p
p
p
? Choose
1
1
1
answer:
\newline
(A)
p
2
+
5
p
−
56
=
0
p^{2}+5p-56=0
p
2
+
5
p
−
56
=
0
\newline
(B)
p
2
−
5
p
−
36
=
0
p^{2}-5p-36=0
p
2
−
5
p
−
36
=
0
\newline
(C)
p
2
+
5
p
−
36
=
0
p^{2}+5p-36=0
p
2
+
5
p
−
36
=
0
\newline
(D)
p
2
−
5
p
−
56
=
0
p^{2}-5p-56=0
p
2
−
5
p
−
56
=
0
Get tutor help
Find one value of
x
x
x
that is a solution to the equation:
\newline
(
x
2
−
7
)
2
+
2
x
2
−
14
=
0
\left(x^{2}-7\right)^{2}+2x^{2}-14=0
(
x
2
−
7
)
2
+
2
x
2
−
14
=
0
\newline
x
=
□
x=\square
x
=
□
Get tutor help
Find one value of
x
x
x
that is a solution to the equation:
\newline
(
3
x
+
7
)
2
=
−
6
x
−
14
(3x+7)^{2}=-6x-14
(
3
x
+
7
)
2
=
−
6
x
−
14
\newline
x
=
x=
x
=
Get tutor help
Solve for
v
v
v
.
\newline
v
2
+
8
v
+
12
=
0
v^2+8v+12=0
v
2
+
8
v
+
12
=
0
\newline
Write each solution as an integer, proper fraction, or improper fraction in simplest form. If there are multiple solutions, separate them with commas.
\newline
v
=
v=
v
=
_____
Get tutor help
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