Find the Roots (Zeros) 9x^2+38=-36x

$9 x^{2} + 38 = - 36 x$

Add $36 x$ to both sides of the equation.

$9 x^{2} + 38 + 36 x = 0$

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 9$ , $b = 36$ , and $c = 38$ into the quadratic formula and solve for $x$ .

$\frac{- 36 \pm \sqrt{36^{2} - 4 \cdot (9 \cdot 38)}}{2 \cdot 9}$

Simplify.

Simplify the numerator.

Raise $36$ to the power of $2$ .

$x = \frac{- 36 \pm \sqrt{1296 - 4 \cdot 9 \cdot 38}}{2 \cdot 9}$

Multiply $- 4$ by $9$ .

$x = \frac{- 36 \pm \sqrt{1296 - 36 \cdot 38}}{2 \cdot 9}$

Multiply $- 36$ by $38$ .

$x = \frac{- 36 \pm \sqrt{1296 - 1368}}{2 \cdot 9}$

Subtract $1368$ from $1296$ .

$x = \frac{- 36 \pm \sqrt{- 72}}{2 \cdot 9}$

Rewrite $- 72$ as $- 1 (72)$ .

$x = \frac{- 36 \pm \sqrt{- 1 \cdot 72}}{2 \cdot 9}$

Rewrite $\sqrt{- 1 (72)}$ as $\sqrt{- 1} \cdot \sqrt{72}$ .

$x = \frac{- 36 \pm \sqrt{- 1} \cdot \sqrt{72}}{2 \cdot 9}$

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

$x = \frac{- 36 \pm i \cdot \sqrt{72}}{2 \cdot 9}$

Rewrite $72$ as $6^{2} \cdot 2$ .

Factor $36$ out of $72$ .

$x = \frac{- 36 \pm i \cdot \sqrt{36 (2)}}{2 \cdot 9}$

Rewrite $36$ as $6^{2}$ .

$x = \frac{- 36 \pm i \cdot \sqrt{6^{2} \cdot 2}}{2 \cdot 9}$

Pull terms out from under the radical.

$x = \frac{- 36 \pm i \cdot (6 \sqrt{2})}{2 \cdot 9}$

Move $6$ to the left of $i$ .

$x = \frac{- 36 \pm 6 i \sqrt{2}}{2 \cdot 9}$

Multiply $2$ by $9$ .

$x = \frac{- 36 \pm 6 i \sqrt{2}}{18}$

Simplify $\frac{- 36 \pm 6 i \sqrt{2}}{18}$ .

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