Find All Complex Solutions 5x^2+20=16x

$5 x^{2} + 20 = 16 x$

Step 1

Subtract $16 x$ from both sides of the equation.

$5 x^{2} + 20 - 16 x = 0$

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 $a = 5$ , $b = - 16$ , and $c = 20$ into the quadratic formula and solve for $x$ .

$\frac{16 \pm \sqrt{{(- 16)}^{2} - 4 \cdot (5 \cdot 20)}}{2 \cdot 5}$

Step 4

Simplify.

Simplify the numerator.

Raise $- 16$ to the power of $2$ .

$x = \frac{16 \pm \sqrt{256 - 4 \cdot 5 \cdot 20}}{2 \cdot 5}$

Multiply $- 4 \cdot 5 \cdot 20$ .

Multiply $- 4$ by $5$ .

$x = \frac{16 \pm \sqrt{256 - 20 \cdot 20}}{2 \cdot 5}$

Multiply $- 20$ by $20$ .

$x = \frac{16 \pm \sqrt{256 - 400}}{2 \cdot 5}$

Subtract $400$ from $256$ .

$x = \frac{16 \pm \sqrt{- 144}}{2 \cdot 5}$

Rewrite $- 144$ as $- 1 (144)$ .

$x = \frac{16 \pm \sqrt{- 1 \cdot 144}}{2 \cdot 5}$

Rewrite $\sqrt{- 1 (144)}$ as $\sqrt{- 1} \cdot \sqrt{144}$ .

$x = \frac{16 \pm \sqrt{- 1} \cdot \sqrt{144}}{2 \cdot 5}$

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

$x = \frac{16 \pm i \cdot \sqrt{144}}{2 \cdot 5}$

Rewrite $144$ as $12^{2}$ .

$x = \frac{16 \pm i \cdot \sqrt{12^{2}}}{2 \cdot 5}$

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

$x = \frac{16 \pm i \cdot 12}{2 \cdot 5}$

Move $12$ to the left of $i$ .

$x = \frac{16 \pm 12 i}{2 \cdot 5}$

Multiply $2$ by $5$ .

$x = \frac{16 \pm 12 i}{10}$

Simplify $\frac{16 \pm 12 i}{10}$ .

$x = \frac{8 \pm 6 i}{5}$