Find All Complex Solutions x^2+10x=-89

$x^{2} + 10 x = - 89$

Step 1

Add $89$ to both sides of the equation.

$x^{2} + 10 x + 89 = 0$

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 $a = 1$ , $b = 10$ , and $c = 89$ into the quadratic formula and solve for $x$ .

$\frac{- 10 \pm \sqrt{10^{2} - 4 \cdot (1 \cdot 89)}}{2 \cdot 1}$

Step 4

Simplify.

Simplify the numerator.

Raise $10$ to the power of $2$ .

$x = \frac{- 10 \pm \sqrt{100 - 4 \cdot 1 \cdot 89}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 89$ .

Multiply $- 4$ by $1$ .

$x = \frac{- 10 \pm \sqrt{100 - 4 \cdot 89}}{2 \cdot 1}$

Multiply $- 4$ by $89$ .

$x = \frac{- 10 \pm \sqrt{100 - 356}}{2 \cdot 1}$

Subtract $356$ from $100$ .

$x = \frac{- 10 \pm \sqrt{- 256}}{2 \cdot 1}$

Rewrite $- 256$ as $- 1 (256)$ .

$x = \frac{- 10 \pm \sqrt{- 1 \cdot 256}}{2 \cdot 1}$

Rewrite $\sqrt{- 1 (256)}$ as $\sqrt{- 1} \cdot \sqrt{256}$ .

$x = \frac{- 10 \pm \sqrt{- 1} \cdot \sqrt{256}}{2 \cdot 1}$

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

$x = \frac{- 10 \pm i \cdot \sqrt{256}}{2 \cdot 1}$

Rewrite $256$ as $16^{2}$ .

$x = \frac{- 10 \pm i \cdot \sqrt{16^{2}}}{2 \cdot 1}$

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

$x = \frac{- 10 \pm i \cdot 16}{2 \cdot 1}$

Move $16$ to the left of $i$ .

$x = \frac{- 10 \pm 16 i}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 10 \pm 16 i}{2}$

Simplify $\frac{- 10 \pm 16 i}{2}$ .

$x = - 5 \pm 8 i$