Solve for q q^2+20q=-104

$q^{2} + 20 q = - 104$

Add $104$ to both sides of the equation.

$q^{2} + 20 q + 104 = 0$

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 1$ , $b = 20$ , and $c = 104$ into the quadratic formula and solve for $q$ .

$\frac{- 20 \pm \sqrt{20^{2} - 4 \cdot (1 \cdot 104)}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

Raise $20$ to the power of $2$ .

$q = \frac{- 20 \pm \sqrt{400 - 4 \cdot 1 \cdot 104}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$q = \frac{- 20 \pm \sqrt{400 - 4 \cdot 104}}{2 \cdot 1}$

Multiply $- 4$ by $104$ .

$q = \frac{- 20 \pm \sqrt{400 - 416}}{2 \cdot 1}$

Subtract $416$ from $400$ .

$q = \frac{- 20 \pm \sqrt{- 16}}{2 \cdot 1}$

Rewrite $- 16$ as $- 1 (16)$ .

$q = \frac{- 20 \pm \sqrt{- 1 \cdot 16}}{2 \cdot 1}$

Rewrite $\sqrt{- 1 (16)}$ as $\sqrt{- 1} \cdot \sqrt{16}$ .

$q = \frac{- 20 \pm \sqrt{- 1} \cdot \sqrt{16}}{2 \cdot 1}$

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

$q = \frac{- 20 \pm i \cdot \sqrt{16}}{2 \cdot 1}$

Rewrite $16$ as $4^{2}$ .

$q = \frac{- 20 \pm i \cdot \sqrt{4^{2}}}{2 \cdot 1}$

Pull terms out from under the radical, assuming positive real numbers.

$q = \frac{- 20 \pm i \cdot 4}{2 \cdot 1}$

Move $4$ to the left of $i$ .

$q = \frac{- 20 \pm 4 i}{2 \cdot 1}$

Multiply $2$ by $1$ .

$q = \frac{- 20 \pm 4 i}{2}$

Simplify $\frac{- 20 \pm 4 i}{2}$ .

$q = - 10 \pm 2 i$

The final answer is the combination of both solutions.

$q = - 10 + 2 i, - 10 - 2 i$