Find the Roots (Zeros) 2x^2-10x=-7x-2

$2 x^{2} - 10 x = - 7 x - 2$

Move all terms containing $x$ to the left side of the equation.

Add $7 x$ to both sides of the equation.

$2 x^{2} - 10 x + 7 x = - 2$

Add $- 10 x$ and $7 x$ .

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

Add $2$ to both sides of the equation.

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

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 2$ , $b = - 3$ , and $c = 2$ into the quadratic formula and solve for $x$ .

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

Simplify.

Simplify the numerator.

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

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

Multiply $- 4$ by $2$ .

$x = \frac{3 \pm \sqrt{9 - 8 \cdot 2}}{2 \cdot 2}$

Multiply $- 8$ by $2$ .

$x = \frac{3 \pm \sqrt{9 - 16}}{2 \cdot 2}$

Subtract $16$ from $9$ .

$x = \frac{3 \pm \sqrt{- 7}}{2 \cdot 2}$

Rewrite $- 7$ as $- 1 (7)$ .

$x = \frac{3 \pm \sqrt{- 1 \cdot 7}}{2 \cdot 2}$

Rewrite $\sqrt{- 1 (7)}$ as $\sqrt{- 1} \cdot \sqrt{7}$ .

$x = \frac{3 \pm \sqrt{- 1} \cdot \sqrt{7}}{2 \cdot 2}$

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{3 \pm i \sqrt{7}}{2 \cdot 2}$

Multiply $2$ by $2$ .

$x = \frac{3 \pm i \sqrt{7}}{4}$