Answer
$x\approx-0.81
\text{ and }
x\approx0.31$
Work Step by Step
Using the properties of equality, the given equation, $
2x^2+x=\dfrac{1}{2}
,$ is equivalent to
\begin{align*}
2(2x^2+x)&=\left(\dfrac{1}{2}\right)2
\\\\
4x^2+2x&=1
\\
4x^2+2x-1&=0
\end{align*}
In the equation above, $a=
4
,$ $b=
2
,$ and $c=
-1
.$
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(4)(-1)}}{2(4)}
\\\\&=
\dfrac{-2\pm\sqrt{4+16}}{8}
\\\\&=
\dfrac{-2\pm\sqrt{20}}{8}
\\\\&=
\dfrac{-2\pm\sqrt{4\cdot5}}{8}
\\\\&=
\dfrac{-2\pm2\sqrt{5}}{8}
\\\\&=
\dfrac{-\cancel2^1\pm\cancel2^1\sqrt{5}}{\cancel8^4}
&\text{ (divide by $2$)}
\\\\&=
\dfrac{-1\pm\sqrt{5}}{4}
\end{align*}
\begin{array}{lcl}
&\Rightarrow
\dfrac{-1-\sqrt{5}}{4} &\text{ OR }& \dfrac{-1+\sqrt{5}}{4}
\\\\&
\approx-0.81 && \approx0.31
\end{array}
Hence, the solutions are $
x\approx -0.81
\text{ and }
x\approx 0.31
.$
