Solve Using the Quadratic Formula 6+2x^2-3x=8x^2

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

Step 1

Move all terms to the left side of the equation and simplify.

Subtract $8 x^{2}$ from both sides of the equation.

$6 + 2 x^{2} - 3 x - 8 x^{2} = 0$

Subtract $8 x^{2}$ from $2 x^{2}$ .

$6 - 6 x^{2} - 3 x = 0$

Step 2

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 3

Substitute the values $a = - 6$ , $b = - 3$ , and $c = 6$ into the quadratic formula and solve for $x$ .

$\frac{3 \pm \sqrt{{(- 3)}^{2} - 4 \cdot (- 6 \cdot 6)}}{2 \cdot - 6}$

Step 4

Simplify.

Simplify the numerator.

Raise $- 3$ to the power of $2$ .

$x = \frac{3 \pm \sqrt{9 - 4 \cdot - 6 \cdot 6}}{2 \cdot - 6}$

Multiply $- 4 \cdot - 6 \cdot 6$ .

Multiply $- 4$ by $- 6$ .

$x = \frac{3 \pm \sqrt{9 + 24 \cdot 6}}{2 \cdot - 6}$

Multiply $24$ by $6$ .

$x = \frac{3 \pm \sqrt{9 + 144}}{2 \cdot - 6}$

Add $9$ and $144$ .

$x = \frac{3 \pm \sqrt{153}}{2 \cdot - 6}$

Rewrite $153$ as $3^{2} \cdot 17$ .

Factor $9$ out of $153$ .

$x = \frac{3 \pm \sqrt{9 (17)}}{2 \cdot - 6}$

Rewrite $9$ as $3^{2}$ .

$x = \frac{3 \pm \sqrt{3^{2} \cdot 17}}{2 \cdot - 6}$

Pull terms out from under the radical.

$x = \frac{3 \pm 3 \sqrt{17}}{2 \cdot - 6}$

Multiply $2$ by $- 6$ .

$x = \frac{3 \pm 3 \sqrt{17}}{- 12}$

Simplify $\frac{3 \pm 3 \sqrt{17}}{- 12}$ .

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

Move the negative in front of the fraction.

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

Step 5

The result can be shown in multiple forms.

Exact Form:

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

Decimal Form:

$x = - 1.28077640 \dots, 0.78077640 \dots$