Solve Using the Quadratic Formula 5y^2-y-4=0

$5 y^{2} - y - 4 = 0$

Step 1

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 2

Substitute the values $a = 5$ , $b = - 1$ , and $c = - 4$ into the quadratic formula and solve for $y$ .

$\frac{1 \pm \sqrt{{(- 1)}^{2} - 4 \cdot (5 \cdot - 4)}}{2 \cdot 5}$

Step 3

Simplify.

Simplify the numerator.

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

$y = \frac{1 \pm \sqrt{1 - 4 \cdot 5 \cdot - 4}}{2 \cdot 5}$

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

Multiply $- 4$ by $5$ .

$y = \frac{1 \pm \sqrt{1 - 20 \cdot - 4}}{2 \cdot 5}$

Multiply $- 20$ by $- 4$ .

$y = \frac{1 \pm \sqrt{1 + 80}}{2 \cdot 5}$

Add $1$ and $80$ .

$y = \frac{1 \pm \sqrt{81}}{2 \cdot 5}$

Rewrite $81$ as $9^{2}$ .

$y = \frac{1 \pm \sqrt{9^{2}}}{2 \cdot 5}$

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

$y = \frac{1 \pm 9}{2 \cdot 5}$

Multiply $2$ by $5$ .

$y = \frac{1 \pm 9}{10}$

Step 4

The final answer is the combination of both solutions.

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