Answer
the solution set is\[\left\{ \frac{4}{3},-1 \right\}\].
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: \[3{{x}^{2}}-x-4=0\]
Step 2: Find the factor of the above equation:
Consider the equation, \[3{{x}^{2}}-x-4=0\].
Factorize it as follows:
\[\begin{align}
& 3{{x}^{2}}-x-4=0 \\
& 3{{x}^{2}}-4x+3x-4=0 \\
& x\left( 3x-4 \right)+1\left( 3x-4 \right)=0 \\
& \left( 3x-4 \right)\left( x+1 \right)=0
\end{align}\]
Steps 3 and 4: Set each factor equal to zero and solve the resulting equation:
From step 2, \[\left( 3x-4 \right)\left( x+1 \right)=0\].
By the zero product principal, either \[\left( 3x-4 \right)=0\]or\[\left( x+1 \right)=0\].
Now \[\left( 3x-4 \right)=0\]implies that \[x=\frac{4}{3}\]and \[\left( x+1 \right)=0\] implies that\[x=-1\].
Step 5: Check the solution in the original equation.
Check for\[x=-1\]. So consider,
\[\begin{align}
& 3{{x}^{2}}-x-4=0 \\
& 3{{\left( -1 \right)}^{2}}-1\left( -1 \right)-4=0 \\
& 3+1-4=0 \\
& 0=0
\end{align}\]
Now check for\[x=\frac{4}{3}\]. So consider,
\[\begin{align}
& 3{{x}^{2}}-x-4=0 \\
& 3{{\left( \frac{4}{3} \right)}^{2}}-1\left( \frac{4}{3} \right)-4=0 \\
& \frac{16}{3}-\frac{4}{3}-4=0 \\
& 0=0
\end{align}\]
Hence, the solution set is\[\left\{ \frac{4}{3},-1 \right\}\].
