Solve for y y^2+4y-4=0

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

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

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

Step 3

Simplify.

Simplify the numerator.

Raise $4$ to the power of $2$ .

$y = \frac{- 4 \pm \sqrt{16 - 4 \cdot 1 \cdot - 4}}{2 \cdot 1}$

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

Multiply $- 4$ by $1$ .

$y = \frac{- 4 \pm \sqrt{16 - 4 \cdot - 4}}{2 \cdot 1}$

Multiply $- 4$ by $- 4$ .

$y = \frac{- 4 \pm \sqrt{16 + 16}}{2 \cdot 1}$

Add $16$ and $16$ .

$y = \frac{- 4 \pm \sqrt{32}}{2 \cdot 1}$

Rewrite $32$ as $4^{2} \cdot 2$ .

Factor $16$ out of $32$ .

$y = \frac{- 4 \pm \sqrt{16 (2)}}{2 \cdot 1}$

Rewrite $16$ as $4^{2}$ .

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

Pull terms out from under the radical.

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

Multiply $2$ by $1$ .

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

Simplify $\frac{- 4 \pm 4 \sqrt{2}}{2}$ .

$y = - 2 \pm 2 \sqrt{2}$

Step 4

The final answer is the combination of both solutions.

$y = - 2 + 2 \sqrt{2}, - 2 - 2 \sqrt{2}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$y = - 2 + 2 \sqrt{2}, - 2 - 2 \sqrt{2}$

Decimal Form:

$y = 0.82842712 \dots, - 4.82842712 \dots$