Answer
The solution set is\[\left\{ -5,-2 \right\}\].
Work Step by Step
Step1:Shift all nonzero terms to left side and obtain zero on the other side and get the equation:
\[3{{x}^{2}}+21x+30=0\]
Step2:Find the factor of the above equation:
Consider the equation,\[3{{x}^{2}}+21x+30=0\]
Factorize it as follows:
\[\begin{align}
& 3{{x}^{2}}+21x+30=0 \\
& 3{{x}^{2}}+15x+6x+30=0 \\
& 3x\left( x+5 \right)+6\left( x+5 \right)=0 \\
& \left( 3x+6 \right)\left( x+5 \right)=0
\end{align}\]
Steps3 and 4: Set each factor equal to zero and solve the resulting equation:
From Step 2, \[\left( 3x+6 \right)\left( x+5 \right)=0\]
By the zero product principal either \[\left( 3x+6 \right)=0\]or\[\left( x+5 \right)=0\].
Now \[\left( 3x+6 \right)=0\] implies that\[x=-\frac{6}{3}=-2\] and\[\left( x+5 \right)=0\]implies that\[x=-5\].
Step5: Check the solution in the original equation:
Check for\[x=-5\]. So consider,
\[\begin{align}
& 3{{x}^{2}}+21x+30=0 \\
& 3{{\left( -5 \right)}^{2}}+21\left( -5 \right)+30=0 \\
& 75-105+30=0 \\
& 0=0
\end{align}\]
and, check for\[x=-2\].
\[\begin{align}
& 3{{x}^{2}}+21x+30=0 \\
& 3{{\left( -2 \right)}^{2}}+21\left( -2 \right)+30=0 \\
& 12-42+30=0 \\
& 0=0
\end{align}\]
