Solve. -7x^(2)+x+9=-6x Choose 1 answer: (A) x=(-7+-sqrt301)/(-14) (B) x=-1,(7)/(10) (C) x=(2+-sqrt5)/(-2) (D) x=(1+-sqrt57)/(-8)

Moving terms to one side: First, we need to move all terms to one side of the equation to set it equal to zero. -7x^2 + x + 9 = -6x Add 6x to both sides to get: -7x^2 + 7x + 9 = 0Identifying coefficients: Now, we identify the coefficients a, b, and c in the quadratic equation ax^2 + bx + c = 0. In our equation, -7x^2 + 7x + 9 = 0, we have: a = -7 b = 7 c = 9Using the quadratic formula: Next, we use the quadratic formula to find the solutions for x: x = [-b ± sqrt(b^2 - 4ac)] / (2a) Substitute a = -7, b = 7, and c = 9 into the formula. x = [-7 ± sqrt(7^2 - 4(-7)(9))] / (2(-7))Calculating the discriminant: Now, we calculate the discriminant (the part under the square root): Discriminant = b^2 - 4ac Discriminant = 7^2 - 4(-7)(9) Discriminant = 49 + 252 Discriminant = 301Substituting the discriminant: We substitute the discriminant back into the quadratic formula: x = [-7 ± sqrt(301)] / (-14)Simplifying the expression: Finally, we simplify the expression: x = (-7 + sqrt(301)) / (-14) or x = (-7 - sqrt(301)) / (-14) These are the solutions in the simplest radical form.

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Solve.

-7x^(2)+x+9=-6x
Choose 1 answer:
(A)
x=(-7+-sqrt301)/(-14)
(B)
x=-1,(7)/(10)
(C)
x=(2+-sqrt5)/(-2)
(D)
x=(1+-sqrt57)/(-8)

Solve. $\newline$ $-7 x^{2}+x+9=-6 x$ $\newline$ Choose $1$ answer: $\newline$ (A) $x=\frac{-7 \pm \sqrt{301}}{-14}$ $\newline$ (B) $x=-1, \frac{7}{10}$ $\newline$ (C) $x=\frac{2 \pm \sqrt{5}}{-2}$ $\newline$ (D) $x=\frac{1 \pm \sqrt{57}}{-8}$

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Math Problems

Algebra 2

Solve a quadratic equation using the quadratic formula

Full solution

Q. Solve.

\newline

-7 x^{2}+x+9=-6 x

\newline

Choose

1

answer:

\newline

(A)

x=\frac{-7 \pm \sqrt{301}}{-14}

\newline

(B)

x=-1, \frac{7}{10}

\newline

(C)

x=\frac{2 \pm \sqrt{5}}{-2}

\newline

(D)

x=\frac{1 \pm \sqrt{57}}{-8}

Moving terms to one side: First, we need to move all terms to one side of the equation to set it equal to zero. $\newline$ $-7x^2 + x + 9 = -6x$ $\newline$ Add $6x$ to both sides to get: $\newline$ $-7x^2 + 7x + 9 = 0$
Identifying coefficients: Now, we identify the coefficients $a$ , $b$ , and $c$ in the quadratic equation $ax^2 + bx + c = 0$ . $\newline$ In our equation, $-7x^2 + 7x + 9 = 0$ , we have: $\newline$ $a = -7$ $\newline$ $b = 7$ $\newline$ $c = 9$
Using the quadratic formula: Next, we use the quadratic formula to find the solutions for $x$ : $x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}$ Substitute $a = -7$ , $b = 7$ , and $c = 9$ into the formula. $x = \frac{{-7 \pm \sqrt{{7^2 - 4(-7)(9)}}}}{{2(-7)}}$
Calculating the discriminant: Now, we calculate the discriminant (the part under the square root): $\newline$ Discriminant = $b^2 - 4ac$ $\newline$ Discriminant = $7^2 - 4(-7)(9)$ $\newline$ Discriminant = $49 + 252$ $\newline$ Discriminant = $301$
Substituting the discriminant: We substitute the discriminant back into the quadratic formula: $\newline$ $x = \frac{{-7 \pm \sqrt{301}}}{{-14}}$
Simplifying the expression: Finally, we simplify the expression: $x = \frac{-7 + \sqrt{301}}{-14}$ or $x = \frac{-7 - \sqrt{301}}{-14}$ These are the solutions in the simplest radical form.