Answer
the solution set is\[\left\{ \frac{-5+\sqrt{13}}{2},\frac{-5-\sqrt{13}}{2} \right\}\].
Work Step by Step
The given equation can be written as \[{{x}^{2}}+5x+3=0\].
Compare the given equation with the equation\[a{{x}^{2}}+bx+c=0\].
Here,\[a=1,\,\,b=5,\text{ and }c=3\]
Now put these values in the quadratic formula: \[x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
That is,
\[\begin{align}
& x=\frac{-5\pm \sqrt{{{\left( 5 \right)}^{2}}-4\times 1\times \left( 3 \right)}}{2\times 1} \\
& =\frac{-5\pm \sqrt{25-12}}{2} \\
& =\frac{-5\pm \sqrt{13}}{2}
\end{align}\]
Further simplify it to get,
\[\begin{align}
& x=\frac{-5\pm \sqrt{13}}{2} \\
& =\frac{-5+\sqrt{13}}{2},\frac{-5-\sqrt{13}}{2}
\end{align}\]
Hence, the solution set is\[\left\{ \frac{-5+\sqrt{13}}{2},\frac{-5-\sqrt{13}}{2} \right\}\].
