Answer
The solution set of equation $2{{x}^{2}}-3x-2=0$ is $\left\{ \left. -\frac{1}{2},2 \right\} \right.$.
Work Step by Step
Consider the provided equation:
$2{{x}^{2}}-3x-2=0$
On factorizing, we get:
$\begin{align}
& 2{{x}^{2}}-4x+x-2=0 \\
& 2x\left( x-2 \right)+1\left( x-2 \right)=0 \\
& \left( x-2 \right)\left( 2x+1 \right)=0
\end{align}$
Now use the principle of zero products:
$\left( x-2 \right)=0$ or $\left( 2x+1 \right)=0$
From here
$\begin{align}
& \left( x-2 \right)=0 \\
& x=2
\end{align}$
Or
$\begin{align}
& \left( 2x+1 \right)=0 \\
& 2x=-1 \\
& x=\frac{-1}{2}
\end{align}$
Check:
For $x=2$
Put $x=2$ in $2{{x}^{2}}-3x-2=0$
$\begin{align}
& 2{{\left( 2 \right)}^{2}}-3\left( 2 \right)-2\overset{?}{\mathop{=}}\,0 \\
& 2\left( 4 \right)-6-2\overset{?}{\mathop{=}}\,0 \\
& 8-8\overset{?}{\mathop{=}}\,0 \\
& 0=0
\end{align}$
For $x=\frac{-1}{2}$
Put $x=\frac{-1}{2}$ in $2{{x}^{2}}-3x-2=0$
$\begin{align}
& 2{{\left( \frac{-1}{2} \right)}^{2}}-3\left( \frac{-1}{2} \right)-2\overset{?}{\mathop{=}}\,0 \\
& 2\left( \frac{1}{4} \right)+\frac{3}{2}-2\overset{?}{\mathop{=}}\,0 \\
& \frac{1}{2}-\frac{1}{2}\overset{?}{\mathop{=}}\,0 \\
& 0=0
\end{align}$
Which is true.
Therefore, the solution set of the equation $2{{x}^{2}}-3x-2=0$ is $\left\{ \left. -\frac{1}{2},2 \right\} \right.$.
