Answer
$x\approx-3.45
\text{ and }
x\approx1.45$
Work Step by Step
Using the properties of equality, the given equation, $
2x^2+4x=10
,$ is equivalent to
\begin{align*}
2x^2+4x-10&=0
\\\\
\dfrac{2x^2+4x-10}{2}&=\dfrac{0}{2}
\\\\
x^2+2x-5&=0
\end{align*}
In the equation above, $a=
1
,$ $b=
2
,$ and $c=
-5
.$
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(1)(-5)}}{2(1)}
\\\\&=
\dfrac{-2\pm\sqrt{4+20}}{2}
\\\\&=
\dfrac{-2\pm\sqrt{24}}{2}
\\\\&=
\dfrac{-2\pm\sqrt{4\cdot6}}{2}
\\\\&=
\dfrac{-2\pm2\sqrt{6}}{2}
\\\\&=
\dfrac{-\cancel2^1\pm\cancel2^1\sqrt{6}}{\cancel2^1}
\\\\&=
\dfrac{-1\pm\sqrt{6}}{1}
\\\\&=
-1\pm\sqrt{6}
\end{align*}
\begin{array}{lcl}
&\Rightarrow
-1-\sqrt{6} &\text{ OR }& -1+\sqrt{6}
\\\\&
\approx-3.45 && \approx1.45
\end{array}
Hence, the solutions are $
x\approx-3.45
\text{ and }
x\approx1.45
.$
