Find the Roots (Zeros) 20x^2+1=x

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

Step 1

Subtract $x$ from both sides of the equation.

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

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

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

Step 4

Simplify.

Simplify the numerator.

Raise $- 1$ to the power of $2$ .

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

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

Multiply $- 4$ by $20$ .

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

Multiply $- 80$ by $1$ .

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

Subtract $80$ from $1$ .

$x = \frac{1 \pm \sqrt{- 79}}{2 \cdot 20}$

Rewrite $- 79$ as $- 1 (79)$ .

$x = \frac{1 \pm \sqrt{- 1 \cdot 79}}{2 \cdot 20}$

Rewrite $\sqrt{- 1 (79)}$ as $\sqrt{- 1} \cdot \sqrt{79}$ .

$x = \frac{1 \pm \sqrt{- 1} \cdot \sqrt{79}}{2 \cdot 20}$

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

$x = \frac{1 \pm i \sqrt{79}}{2 \cdot 20}$

Multiply $2$ by $20$ .

$x = \frac{1 \pm i \sqrt{79}}{40}$

Step 5