Solve Using the Quadratic Formula 4y(y+4)=20

$4 y (y + 4) = 20$

Step 1

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

Simplify the left side.

Simplify $4 y (y + 4)$ .

Apply the distributive property.

$4 y \cdot y + 4 y \cdot 4 = 20$

Multiply $y$ by $y$ by adding the exponents.

Move $y$ .

$4 (y \cdot y) + 4 y \cdot 4 = 20$

Multiply $y$ by $y$ .

$4 y^{2} + 4 y \cdot 4 = 20$

Multiply $4$ by $4$ .

$4 y^{2} + 16 y = 20$

Subtract $20$ from both sides of the equation.

$4 y^{2} + 16 y - 20 = 0$

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 $a = 4$ , $b = 16$ , and $c = - 20$ into the quadratic formula and solve for $y$ .

$\frac{- 16 \pm \sqrt{16^{2} - 4 \cdot (4 \cdot - 20)}}{2 \cdot 4}$

Step 4

Simplify.

Simplify the numerator.

Raise $16$ to the power of $2$ .

$y = \frac{- 16 \pm \sqrt{256 - 4 \cdot 4 \cdot - 20}}{2 \cdot 4}$

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

Multiply $- 4$ by $4$ .

$y = \frac{- 16 \pm \sqrt{256 - 16 \cdot - 20}}{2 \cdot 4}$

Multiply $- 16$ by $- 20$ .

$y = \frac{- 16 \pm \sqrt{256 + 320}}{2 \cdot 4}$

Add $256$ and $320$ .

$y = \frac{- 16 \pm \sqrt{576}}{2 \cdot 4}$

Rewrite $576$ as $24^{2}$ .

$y = \frac{- 16 \pm \sqrt{24^{2}}}{2 \cdot 4}$

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

$y = \frac{- 16 \pm 24}{2 \cdot 4}$

Multiply $2$ by $4$ .

$y = \frac{- 16 \pm 24}{8}$

Simplify $\frac{- 16 \pm 24}{8}$ .

$y = - 2 \pm 3$

Step 5

The final answer is the combination of both solutions.

$y = 1, - 5$