Answer
the solution set is\[\left\{ -\frac{1}{2},4 \right\}\].
Work Step by Step
Step 1: Shift all nonzero terms to left side and obtain zero on the other side.
After shifting all nonzero terms to the left side, the above equation becomes: \[2{{x}^{2}}-7x-4=0\]
Step 2: Find the factor of the above equation:
Consider the equation, \[2{{x}^{2}}-7x-4=0\].
Factorize it as follows:
\[\begin{align}
& 2{{x}^{2}}-7x-4=0 \\
& 2{{x}^{2}}-8x+x-4=0 \\
& 2x\left( x-4 \right)+1\left( x-4 \right)=0 \\
& \left( 2x+1 \right)\left( x-4 \right)=0
\end{align}\]
Steps 3 and 4: Set each factor equal to zero and solve the resulting equation:
From step 2, \[\left( 2x+1 \right)\left( x-4 \right)=0\]
By the zero product principal, either \[\left( 2x+1 \right)=0\]or\[\left( x-4 \right)=0\].
Now \[\left( 2x+1 \right)=0\]implies that \[x=-\frac{1}{2}\]and \[\left( x-4 \right)=0\] implies that\[x=4\].
Step 5: Check the solution in the original equation.
Check for\[x=4\]. So consider,
\[\begin{align}
& 2{{x}^{2}}-7x-4=0 \\
& 2{{\left( 4 \right)}^{2}}-7\left( 4 \right)-4=0 \\
& 32-28-4=0 \\
& 0=0
\end{align}\]
Now check for\[x=-\frac{1}{2}\]. So consider,
\[\begin{align}
& 2{{x}^{2}}-7x-4=0 \\
& 2{{\left( -\frac{1}{2} \right)}^{2}}-7\left( -\frac{1}{2} \right)-4=0 \\
& \frac{1}{2}+\frac{7}{2}-4=0 \\
& 0=0
\end{align}\]
Hence, the solution set is\[\left\{ -\frac{1}{2},4 \right\}\].
