Solve Using the Quadratic Formula -8x+10-3x^2=-12x

$- 8 x + 10 - 3 x^{2} = - 12 x$

Step 1

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

Add $12 x$ to both sides of the equation.

$- 8 x + 10 - 3 x^{2} + 12 x = 0$

Add $- 8 x$ and $12 x$ .

$4 x + 10 - 3 x^{2} = 0$

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 $a = - 3$ , $b = 4$ , and $c = 10$ into the quadratic formula and solve for $x$ .

$\frac{- 4 \pm \sqrt{4^{2} - 4 \cdot (- 3 \cdot 10)}}{2 \cdot - 3}$

Step 4

Simplify.

Simplify the numerator.

Raise $4$ to the power of $2$ .

$x = \frac{- 4 \pm \sqrt{16 - 4 \cdot - 3 \cdot 10}}{2 \cdot - 3}$

Multiply $- 4 \cdot - 3 \cdot 10$ .

Multiply $- 4$ by $- 3$ .

$x = \frac{- 4 \pm \sqrt{16 + 12 \cdot 10}}{2 \cdot - 3}$

Multiply $12$ by $10$ .

$x = \frac{- 4 \pm \sqrt{16 + 120}}{2 \cdot - 3}$

Add $16$ and $120$ .

$x = \frac{- 4 \pm \sqrt{136}}{2 \cdot - 3}$

Rewrite $136$ as $2^{2} \cdot 34$ .

Factor $4$ out of $136$ .

$x = \frac{- 4 \pm \sqrt{4 (34)}}{2 \cdot - 3}$

Rewrite $4$ as $2^{2}$ .

$x = \frac{- 4 \pm \sqrt{2^{2} \cdot 34}}{2 \cdot - 3}$

Pull terms out from under the radical.

$x = \frac{- 4 \pm 2 \sqrt{34}}{2 \cdot - 3}$

Multiply $2$ by $- 3$ .

$x = \frac{- 4 \pm 2 \sqrt{34}}{- 6}$

Simplify $\frac{- 4 \pm 2 \sqrt{34}}{- 6}$ .

$x = \frac{2 \pm \sqrt{34}}{3}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$x = \frac{2 \pm \sqrt{34}}{3}$

Decimal Form:

$x = 2.61031729 \dots, - 1.27698396 \dots$