Answer
the solution set is\[\left\{ -2,\frac{9}{5} \right\}\].
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: \[5{{x}^{2}}+x-18=0\]
Step 2: Find the factor of the above equation:
Consider the equation \[5{{x}^{2}}+x-18=0\].
Factorize it as follows:
\[\begin{align}
& 5{{x}^{2}}+x-18=0 \\
& 5{{x}^{2}}+10x-9x-18=0 \\
& 5x\left( x+2 \right)-9\left( x+2 \right)=0 \\
& \left( 5x-9 \right)\left( x+2 \right)=0
\end{align}\]
Steps 3 and 4: Set each factor equal to zero and solve the resulting equation:
From step 2, \[\left( 5x-9 \right)\left( x+2 \right)=0\].
By the zero product principal, either \[\left( 5x-9 \right)=0\]or\[\left( x+2 \right)=0\].
Now \[\left( 5x-9 \right)=0\]implies that \[x=\frac{9}{5}\]and \[\left( x+2 \right)=0\] implies that\[x=-2\].
Step 5: Check the solution in the original equation.
Check for\[x=-2\]. So consider,
\[\begin{align}
& 5{{x}^{2}}+x-18=0 \\
& 5{{\left( -2 \right)}^{2}}+1\left( -2 \right)-18=0 \\
& 20-2-18=0 \\
& 0=0
\end{align}\]
Now check for\[x=\frac{9}{5}\]. So consider,
\[\begin{align}
& 5{{x}^{2}}+x-18=0 \\
& 5{{\left( \frac{9}{5} \right)}^{2}}+1\left( \frac{9}{5} \right)-18=0 \\
& \frac{81}{5}+\frac{9}{5}-18=0 \\
& 0=0
\end{align}\]
Hence, the solution set is\[\left\{ -2,\frac{9}{5} \right\}\].
