Find All Complex Solutions 20x^2+1=4x

$20 x^{2} + 1 = 4 x$

Step 1

Subtract $4 x$ from both sides of the equation.

$20 x^{2} + 1 - 4 x = 0$

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 $a = 20$ , $b = - 4$ , and $c = 1$ into the quadratic formula and solve for $x$ .

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

Step 4

Simplify.

Simplify the numerator.

Raise $- 4$ to the power of $2$ .

$x = \frac{4 \pm \sqrt{16 - 4 \cdot 20 \cdot 1}}{2 \cdot 20}$

Multiply $- 4 \cdot 20 \cdot 1$ .

Multiply $- 4$ by $20$ .

$x = \frac{4 \pm \sqrt{16 - 80 \cdot 1}}{2 \cdot 20}$

Multiply $- 80$ by $1$ .

$x = \frac{4 \pm \sqrt{16 - 80}}{2 \cdot 20}$

Subtract $80$ from $16$ .

$x = \frac{4 \pm \sqrt{- 64}}{2 \cdot 20}$

Rewrite $- 64$ as $- 1 (64)$ .

$x = \frac{4 \pm \sqrt{- 1 \cdot 64}}{2 \cdot 20}$

Rewrite $\sqrt{- 1 (64)}$ as $\sqrt{- 1} \cdot \sqrt{64}$ .

$x = \frac{4 \pm \sqrt{- 1} \cdot \sqrt{64}}{2 \cdot 20}$

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

$x = \frac{4 \pm i \cdot \sqrt{64}}{2 \cdot 20}$

Rewrite $64$ as $8^{2}$ .

$x = \frac{4 \pm i \cdot \sqrt{8^{2}}}{2 \cdot 20}$

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

$x = \frac{4 \pm i \cdot 8}{2 \cdot 20}$

Move $8$ to the left of $i$ .

$x = \frac{4 \pm 8 i}{2 \cdot 20}$

Multiply $2$ by $20$ .

$x = \frac{4 \pm 8 i}{40}$

Simplify $\frac{4 \pm 8 i}{40}$ .

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