Answer
$x\approx-2.90
\text{ and }
x\approx1.90
$
Work Step by Step
Using the properties of equality, the given equation, $
4x^2+4x=22
,$ is equivalent to
\begin{align*}
4x^2+4x-22&=0
\\\\
\dfrac{4x^2+4x-22}{2}&=\dfrac{0}{2}
\\\\
2x^2+2x-11&=0
\end{align*}
In the equation above, $a=
2
,$ $b=
2
,$ and $c=
-11
.$
Using the Quadratic Formula which is given by $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a},$ then
\begin{align*}\require{cancel}x&=
\dfrac{-2\pm\sqrt{2^2-4(2)(-11)}}{2(2)}
\\\\&=
\dfrac{-2\pm\sqrt{4+88}}{4}
\\\\&=
\dfrac{-2\pm\sqrt{92}}{4}
\\\\&=
\dfrac{-2\pm\sqrt{4\cdot23}}{4}
\\\\&=
\dfrac{-2\pm2\sqrt{23}}{4}
\\\\&=
\dfrac{-\cancel2^1\pm\cancel2^1\sqrt{23}}{\cancel4^2}
&\text{ (divide by $2$)}
\\\\&=
\dfrac{-1\pm\sqrt{23}}{2}
\end{align*}
\begin{array}{lcl}
&\Rightarrow
\dfrac{-1-\sqrt{23}}{2} &\text{ OR }& \dfrac{-1+\sqrt{23}}{2}
\\\\&
\approx-2.90 && \approx1.90
\end{array}
Hence, the solutions are$
x\approx-2.90
\text{ and }
x\approx1.90
.$
