Answer
$x=\left\{
-1-\sqrt{3}, -1+\sqrt{3}
\right\}$
Work Step by Step
Using $ax^2+bx+c=0,$ the given equation,
\begin{align*}\require{cancel}
2x^2+4x-4=0
,\end{align*}
the values of $a,b, \text{ and }c $ are $a=
2
,$ $b=
4
,$ and $c=
-4
.$ Using the Quadratic Formula which is given by $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a},$ then
\begin{align*}
x&=\dfrac{-4\pm\sqrt{4^2-4(2)(-4)}}{2(2)}
\\\\&=
\dfrac{-4\pm\sqrt{16+32}}{4}
\\\\&=
\dfrac{-4\pm\sqrt{48}}{4}
\\\\&=
\dfrac{-4\pm\sqrt{16\cdot3}}{4}
\\\\&=
\dfrac{-4\pm\sqrt{4^2\cdot3}}{4}
\\\\&=
\dfrac{-4\pm4\sqrt{3}}{4}
\\\\&=
\dfrac{-\cancel4^1\pm\cancel4^1\sqrt{3}}{\cancel4^1}
&\text{ (divide by $4$)}
\\\\&=
-1\pm\sqrt{3}
.\end{align*}
The solutions are $
x=\left\{
-1-\sqrt{3}, -1+\sqrt{3}
\right\}
.$
