Find All Complex Solutions 20y^2-20y-5=y

$20 y^{2} - 20 y - 5 = y$

Step 1

Move all terms to the left side of the equation and simplify.

Subtract $y$ from both sides of the equation.

$20 y^{2} - 20 y - 5 - y = 0$

Subtract $y$ from $- 20 y$ .

$20 y^{2} - 21 y - 5 = 0$

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 $a = 20$ , $b = - 21$ , and $c = - 5$ into the quadratic formula and solve for $y$ .

$\frac{21 \pm \sqrt{{(- 21)}^{2} - 4 \cdot (20 \cdot - 5)}}{2 \cdot 20}$

Step 4

Simplify.

Simplify the numerator.

Raise $- 21$ to the power of $2$ .

$y = \frac{21 \pm \sqrt{441 - 4 \cdot 20 \cdot - 5}}{2 \cdot 20}$

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

Multiply $- 4$ by $20$ .

$y = \frac{21 \pm \sqrt{441 - 80 \cdot - 5}}{2 \cdot 20}$

Multiply $- 80$ by $- 5$ .

$y = \frac{21 \pm \sqrt{441 + 400}}{2 \cdot 20}$

Add $441$ and $400$ .

$y = \frac{21 \pm \sqrt{841}}{2 \cdot 20}$

Rewrite $841$ as $29^{2}$ .

$y = \frac{21 \pm \sqrt{29^{2}}}{2 \cdot 20}$

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

$y = \frac{21 \pm 29}{2 \cdot 20}$

Multiply $2$ by $20$ .

$y = \frac{21 \pm 29}{40}$

Step 5

The final answer is the combination of both solutions.

$y = \frac{5}{4}, - \frac{1}{5}$