Answer
the solution set is\[\left\{ -3,-5 \right\}\]
Work Step by Step
The given equation can be written as \[{{x}^{2}}+8x+15=0\].
Compare the given equation with the equation\[a{{x}^{2}}+bx+c=0\].
Here,\[a=1,\,\,b=8,\text{ and }c=15\]
Now put these values in the quadratic formula: \[x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
That is,
\[\begin{align}
& x=\frac{-8\pm \sqrt{{{\left( 8 \right)}^{2}}-4\times 1\times \left( 15 \right)}}{2\times 1} \\
& =\frac{-8\pm \sqrt{64-60}}{2} \\
& =\frac{-8\pm \sqrt{4}}{2} \\
& =\frac{-8\pm 2}{2}
\end{align}\]
Further simplify it to get,
\[\begin{align}
& x=\frac{-8\pm 2}{2} \\
& =\frac{-8+2}{2},\frac{-8-2}{2} \\
& =-\frac{6}{2},-\frac{10}{2} \\
& =-3,-5
\end{align}\]
Hence, the solution set is\[\left\{ -3,-5 \right\}\].
