Solve. -6x-1+5x^(2)=8x^(2) Choose 1 answer: (A) x=(5+-sqrt67)/(6) (B) x=1,(7)/(2) (C) x=(7+-3sqrt41)/(-20) (D) x=(3+-sqrt6)/(-3)

Rearrange equation and set equal to zero: First, we need to rearrange the equation to have all terms on one side and set it equal to zero. -6x - 1 + 5x^2 = 8x^2 Subtract 5x^2 from both sides to get: -6x - 1 = 3x^2 Now, rearrange the terms to get a standard quadratic equation form: 3x^2 + 6x + 1 = 0Use quadratic formula to solve for x: Next, we will use the quadratic formula to solve for x, which is x = (-b ± sqrt(b^2 - 4ac)) / (2a). In our equation, a = 3, b = 6, and c = 1.Substitute values into quadratic formula: Now, we substitute the values of a, b, and c into the quadratic formula: x = (-(6) ± sqrt((6)^2 - 4(3)(1))) / (2(3)) x = (-6 ± sqrt(36 - 12)) / 6 x = (-6 ± sqrt(24)) / 6Simplify square root and equation: Simplify the square root and the equation: x = (-6 ± sqrt(24)) / 6 x = (-6 ± 2sqrt(6)) / 6 Now, we can factor out a 2 from the numerator: x = 2(-3 ± sqrt(6)) / 6Factor out a 2 and simplify fraction: Finally, we simplify the fraction by dividing both the numerator and the denominator by 2: x = (-3 ± sqrt(6)) / 3

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Solve.

-6x-1+5x^(2)=8x^(2)
Choose 1 answer:
(A)
x=(5+-sqrt67)/(6)
(B)
x=1,(7)/(2)
(C)
x=(7+-3sqrt41)/(-20)
(D)
x=(3+-sqrt6)/(-3)

Solve. $\newline$ $-6 x-1+5 x^{2}=8 x^{2}$ $\newline$ Choose $1$ answer: $\newline$ (A) $x=\frac{5 \pm \sqrt{67}}{6}$ $\newline$ (B) $x=1, \frac{7}{2}$ $\newline$ (C) $x=\frac{7 \pm 3 \sqrt{41}}{-20}$ $\newline$ (D) $x=\frac{3 \pm \sqrt{6}}{-3}$

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Math Problems

Algebra 2

Solve a quadratic equation using the quadratic formula

Full solution

Q. Solve.

\newline

-6 x-1+5 x^{2}=8 x^{2}

\newline

Choose

1

answer:

\newline

(A)

x=\frac{5 \pm \sqrt{67}}{6}

\newline

(B)

x=1, \frac{7}{2}

\newline

(C)

x=\frac{7 \pm 3 \sqrt{41}}{-20}

\newline

(D)

x=\frac{3 \pm \sqrt{6}}{-3}

Rearrange equation and set equal to zero: First, we need to rearrange the equation to have all terms on one side and set it equal to zero. $\newline$ $-6x - 1 + 5x^2 = 8x^2$ $\newline$ Subtract $5x^2$ from both sides to get: $\newline$ $-6x - 1 = 3x^2$ $\newline$ Now, rearrange the terms to get a standard quadratic equation form: $\newline$ $3x^2 + 6x + 1 = 0$
Use quadratic formula to solve for x: Next, we will use the quadratic formula to solve for x, which is $x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}$ . $\newline$ In our equation, $a = 3$ , $b = 6$ , and $c = 1$ .
Substitute values into quadratic formula: Now, we substitute the values of $a$ , $b$ , and $c$ into the quadratic formula: $\newline$ $x = \frac{{-\left(6\right) \pm \sqrt{{\left(6\right)^2 - 4(3)(1)}}}}{{2(3)}}$ $\newline$ $x = \frac{{-6 \pm \sqrt{36 - 12}}}{{6}}$ $\newline$ $x = \frac{{-6 \pm \sqrt{24}}}{{6}}$
Simplify square root and equation: Simplify the square root and the equation: $\newline$ $x = \frac{{-6 \pm \sqrt{24}}}{{6}}$ $\newline$ $x = \frac{{-6 \pm 2\sqrt{6}}}{{6}}$ $\newline$ Now, we can factor out a $2$ from the numerator: $\newline$ $x = \frac{{2(-3 \pm \sqrt{6})}}{{6}}$
Factor out a $2$ and simplify fraction: Finally, we simplify the fraction by dividing both the numerator and the denominator by $2$ : $\newline$ $x = \frac{{-3 \pm \sqrt{6}}}{{3}}$