Solve Using the Quadratic Formula -5y^2+2y=-2

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

Step 1

Add $2$ to both sides of the equation.

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

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 $a = - 5$ , $b = 2$ , and $c = 2$ into the quadratic formula and solve for $y$ .

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

Step 4

Simplify.

Simplify the numerator.

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

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

Multiply $- 4 \cdot - 5 \cdot 2$ .

Multiply $- 4$ by $- 5$ .

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

Multiply $20$ by $2$ .

$y = \frac{- 2 \pm \sqrt{4 + 40}}{2 \cdot - 5}$

Add $4$ and $40$ .

$y = \frac{- 2 \pm \sqrt{44}}{2 \cdot - 5}$

Rewrite $44$ as $2^{2} \cdot 11$ .

Factor $4$ out of $44$ .

$y = \frac{- 2 \pm \sqrt{4 (11)}}{2 \cdot - 5}$

Rewrite $4$ as $2^{2}$ .

$y = \frac{- 2 \pm \sqrt{2^{2} \cdot 11}}{2 \cdot - 5}$

Pull terms out from under the radical.

$y = \frac{- 2 \pm 2 \sqrt{11}}{2 \cdot - 5}$

Multiply $2$ by $- 5$ .

$y = \frac{- 2 \pm 2 \sqrt{11}}{- 10}$

Simplify $\frac{- 2 \pm 2 \sqrt{11}}{- 10}$ .

$y = \frac{1 \pm \sqrt{11}}{5}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$y = \frac{1 \pm \sqrt{11}}{5}$

Decimal Form:

$y = 0.86332495 \dots, - 0.46332495 \dots$