Answer
the solution set is\[\left\{ -\frac{5}{3},3 \right\}\].
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: \[3{{x}^{2}}-4x-15=0\]
Step 2: Find the factor of the above equation:
Consider the equation, \[3{{x}^{2}}-4x-15=0\].
Factorize it as follows:
\[\begin{align}
& 3{{x}^{2}}-4x-15=0 \\
& 3{{x}^{2}}-9x+5x-15=0 \\
& 3x\left( x-3 \right)+5\left( x-3 \right)=0 \\
& \left( 3x+5 \right)\left( x-3 \right)=0
\end{align}\]
Steps 3 and 4: Set each factor equal to zero and solve the resulting equation:
From step 2, \[\left( 3x+5 \right)\left( x-3 \right)=0\]
By the zero product principal, either \[\left( 3x+5 \right)=0\]or\[\left( x-3 \right)=0\].
Now \[\left( 3x+5 \right)=0\]implies that \[x=-\frac{5}{3}\]and \[\left( x-3 \right)=0\] implies that\[x=3\].
Step 5: Check the solution in the original equation:
Check for\[x=3\]. So consider,
\[\begin{align}
& 3{{x}^{2}}-4x-15=0 \\
& 3{{\left( 3 \right)}^{2}}-4\left( 3 \right)-15=0 \\
& 27-12-15=0 \\
& 0=0
\end{align}\]
Now check for\[x=-\frac{5}{3}\]. So consider,
\[\begin{align}
& 3{{x}^{2}}-4x-15=0 \\
& 3{{\left( -\frac{5}{3} \right)}^{2}}-4\left( -\frac{5}{3} \right)-15=0 \\
& \frac{25}{3}+\frac{20}{3}-15=0 \\
& 0=0
\end{align}\]
Hence, the solution set is\[\left\{ -\frac{5}{3},3 \right\}\].
