Find the Roots (Zeros) 2x^2+x+2=0

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

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 2$ , $b = 1$ , and $c = 2$ into the quadratic formula and solve for $x$ .

$\frac{- 1 \pm \sqrt{1^{2} - 4 \cdot (2 \cdot 2)}}{2 \cdot 2}$

Simplify.

Simplify the numerator.

One to any power is one.

$x = \frac{- 1 \pm \sqrt{1 - 4 \cdot 2 \cdot 2}}{2 \cdot 2}$

Multiply $- 4$ by $2$ .

$x = \frac{- 1 \pm \sqrt{1 - 8 \cdot 2}}{2 \cdot 2}$

Multiply $- 8$ by $2$ .

$x = \frac{- 1 \pm \sqrt{1 - 16}}{2 \cdot 2}$

Subtract $16$ from $1$ .

$x = \frac{- 1 \pm \sqrt{- 15}}{2 \cdot 2}$

Rewrite $- 15$ as $- 1 (15)$ .

$x = \frac{- 1 \pm \sqrt{- 1 \cdot 15}}{2 \cdot 2}$

Rewrite $\sqrt{- 1 (15)}$ as $\sqrt{- 1} \cdot \sqrt{15}$ .

$x = \frac{- 1 \pm \sqrt{- 1} \cdot \sqrt{15}}{2 \cdot 2}$

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

$x = \frac{- 1 \pm i \sqrt{15}}{2 \cdot 2}$

Multiply $2$ by $2$ .

$x = \frac{- 1 \pm i \sqrt{15}}{4}$