Answer
The solution set of the equation is\[\left\{ \frac{3}{2},-2 \right\}\]
Work Step by Step
Consider the equation:
$2{{x}^{2}}+x=6$
Re-write the above equation as,
$2{{x}^{2}}+x-6=0$
Now factor:
$\begin{align}
& 2{{x}^{2}}+4x-3x-6=0 \\
& 2x\left( x+2 \right)-3\left( x+2 \right)=0 \\
& \left( 2x-3 \right)\left( x+2 \right)=0
\end{align}$
Then equate each term to zero,
$\begin{align}
& 2x-3=0 \\
& 2x=3 \\
& x=\frac{3}{2}
\end{align}$
And,
$\begin{align}
& x+2=0 \\
& x=-2
\end{align}$
Therefore, the value of $x=\frac{3}{2}\text{ and }x=-2$
Check:
Put $\left( x=\frac{3}{2} \right)$ in the provided equation and verify,
$\begin{align}
& 2{{\left( \frac{3}{2} \right)}^{2}}+\frac{3}{2}\overset{?}{\mathop{=}}\,6 \\
& 2\left( \frac{9}{4} \right)+\frac{3}{2}\overset{?}{\mathop{=}}\,6 \\
& \frac{9}{2}+\frac{3}{2}\overset{?}{\mathop{=}}\,6
\end{align}$
Further, simplify
$\begin{align}
\frac{9}{2}+\frac{3}{2}\overset{?}{\mathop{=}}\,6 & \\
\frac{12}{2}\overset{?}{\mathop{=}}\,6 & \\
6=6 & \\
\end{align}$
This is true.
Check:
Put $\left( x=-2 \right)$ in the provided equation and verify,
$\begin{align}
& 2{{\left( -2 \right)}^{2}}+\left( -2 \right)\overset{?}{\mathop{=}}\,6 \\
& 2\left( 4 \right)-2\overset{?}{\mathop{=}}\,6 \\
& 8-2\overset{?}{\mathop{=}}\,6 \\
& 6=6
\end{align}$
This is also true.
Hence, the solution of the equation is $\left\{ \frac{3}{2},-2 \right\}$.
