Answer
the solution set is\[\left\{ -\frac{2}{3},-\frac{3}{2} \right\}\].
Work Step by Step
Step 1: Shift all nonzero terms to left sides and obtain zero on the other side.
Consider the expression\[x\left( 6x+13 \right)+6=0\].
Apply the distributive law to get,
\[6{{x}^{2}}+13x+6=0\]
Step 2: Find the factor of the above equation:
Consider the equation \[6{{x}^{2}}+13x+6=0\].
Factorize it as follows:
\[\begin{align}
& 6{{x}^{2}}+13x+6=0 \\
& 6{{x}^{2}}+9x+4x+6=0 \\
& 3x\left( 2x+3 \right)+2\left( 2x+3 \right)=0 \\
& \left( 2x+3 \right)\left( 3x+2 \right)=0
\end{align}\]
Steps 3 and 4: Set each factor equal to zero and solve the resulting equation:
From step 2, \[\left( 2x+3 \right)\left( 3x+2 \right)=0\].
By the zero product principal, either \[\left( 2x+3 \right)=0\]or\[\left( 3x+2 \right)=0\].
Now \[\left( 2x+3 \right)=0\]implies that \[x=-\frac{3}{2}\]and \[\left( 3x+2 \right)=0\] implies that\[x=-\frac{2}{3}\].
Step 5: Check the solution in the original equation.
Check for\[x=-\frac{3}{2}\]. So consider,
\[\begin{align}
& 6{{x}^{2}}+13x+6=0 \\
& 6{{\left( -\frac{3}{2} \right)}^{2}}+13\left( -\frac{3}{2} \right)+6=0 \\
& \frac{27}{2}-\frac{39}{2}+6=0 \\
& 0=0
\end{align}\]
Now check for\[x=-\frac{2}{3}\]. So consider,
\[\begin{align}
& 6{{x}^{2}}+13x+6=0 \\
& 6{{\left( -\frac{2}{3} \right)}^{2}}+13\left( -\frac{2}{3} \right)+6=0 \\
& \frac{8}{3}-\frac{26}{3}+6=0 \\
& 0=0
\end{align}\]
Hence, the solution set is\[\left\{ -\frac{2}{3},-\frac{3}{2} \right\}\].
