Answer
The solution set is\[\left\{ -3,5 \right\}\].
Work Step by Step
Step 1:Shift all nonzero terms to left side and obtain zero on the other side.
Since all the nonzero terms of the given expression are already on the left side, so no need to do step one.
Step 2:Find the factor of the above equation:
Consider the equation\[{{x}^{2}}-2x-15=0\].
Factorize it as follows:
\[\begin{align}
& {{x}^{2}}-2x-15=0 \\
& {{x}^{2}}-5x+3x-15=0 \\
& x\left( x-5 \right)+3\left( x-5 \right)=0 \\
& \left( x+3 \right)\left( x-5 \right)=0
\end{align}\]
Steps 3 and 4: Set each factor equal to zero and solve the resulting equation:
From step 2, \[\left( x+3 \right)\left( x-5 \right)=0\]
By the zero product principal, either \[\left( x+3 \right)=0\]or\[\left( x-5 \right)=0\].
Now \[\left( x+3 \right)=0\] implies that\[x=-3\] and \[\left( x-5 \right)=0\]implies that\[x=5\].
Step5: Check the solution in the original equation:
Check for\[x=-3\]. So consider,
\[\begin{align}
& {{x}^{2}}-2x-15=0 \\
& {{\left( -3 \right)}^{2}}-2\left( -3 \right)-15=0 \\
& 9+6-15=0 \\
& 0=0
\end{align}\]
And, check for\[x=5\].
\[\begin{align}
& {{x}^{2}}-2x-15=0 \\
& {{\left( 5 \right)}^{2}}-2\left( 5 \right)-15=0 \\
& 25-10-15=0 \\
& 0=0
\end{align}\]
Hence, the solution set is\[\left\{ -3,5 \right\}\].
