Solve. $\newline$ $6+2 x^{2}-3 x=8 x^{2}$ $\newline$ Choose $1$ answer: $\newline$ (A) $x=3,-\frac{1}{2}$ $\newline$ (B) $x=\frac{5 \pm \sqrt{57}}{16}$ $\newline$ (C) $x=\frac{1 \pm \sqrt{17}}{-4}$ $\newline$ (D) $x=\frac{-4 \pm \sqrt{34}}{3}$

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Algebra 2

Solve a quadratic equation using the quadratic formula

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Q. Solve.

\newline

6+2 x^{2}-3 x=8 x^{2}

\newline

Choose

1

answer:

\newline

(A)

x=3,-\frac{1}{2}

\newline

(B)

x=\frac{5 \pm \sqrt{57}}{16}

\newline

(C)

x=\frac{1 \pm \sqrt{17}}{-4}

\newline

(D)

x=\frac{-4 \pm \sqrt{34}}{3}

Setting the equation to zero: First, we need to set the equation to zero by moving all terms to one side. $\newline$ $6 + 2x^2 - 3x = 8x^2$ $\newline$ Subtract $8x^2$ from both sides to get: $\newline$ $6 - 3x - 6x^2 = 0$ $\newline$ Rearrange the terms to get a standard quadratic equation form: $\newline$ $-6x^2 - 3x + 6 = 0$ $\newline$ Now, we can multiply the entire equation by $-1$ to make the $x^2$ coefficient positive: $\newline$ $6x^2 + 3x - 6 = 0$
Rearranging the terms: Now, we will use the quadratic formula to solve for $x$ . The quadratic formula is: $\newline$ $x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}$ $\newline$ For our equation, $a = 6$ , $b = 3$ , and $c = -6$ .
Using the quadratic formula: Let's calculate the discriminant (the part under the square root in the quadratic formula): $\newline$ Discriminant = $b^2 - 4ac$ $\newline$ Discriminant = $(3)^2 - 4(6)(-6)$ $\newline$ Discriminant = $9 + 144$ $\newline$ Discriminant = $153$
Calculating the discriminant: Now we can substitute $a$ , $b$ , and the discriminant into the quadratic formula: $\newline$ $x = \frac{{-3 \pm \sqrt{153}}}{{2 \cdot 6}}$ $\newline$ $x = \frac{{-3 \pm \sqrt{153}}}{{12}}$
Substituting values into the quadratic formula: We can simplify $\sqrt{153}$ to get the exact solutions: $\newline$ $x = \frac{-3 \pm \sqrt{153}}{12}$ $\newline$ Since $\sqrt{153}$ cannot be simplified to an integer or a simple fraction, we will leave it as is.
Simplifying the square root: Now we have two possible solutions for x: $\newline$ $x = \frac{{-3 + \sqrt{153}}}{{12}}$ or $x = \frac{{-3 - \sqrt{153}}}{{12}}$ $\newline$ These are the exact solutions in their simplest form.