Answer
the solution set is\[\left\{ -3\pm \sqrt{19} \right\}\].
Work Step by Step
Compare the given equation with the equation\[a{{x}^{2}}+bx+c=0\].
Here,\[a=1,\,\,b=6,\text{ and }c=-10\]
Now put these values in the quadratic formula: \[x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
That is,
\[\begin{align}
& x=\frac{-6\pm \sqrt{{{\left( 6 \right)}^{2}}-4\times 1\times \left( -10 \right)}}{2\times 1} \\
& =\frac{-6\pm \sqrt{36+40}}{2} \\
& =\frac{-6\pm \sqrt{76}}{2} \\
& =\frac{-6\pm 2\sqrt{19}}{2}
\end{align}\]
Further simplify it to get,
\[\begin{align}
& x=\frac{-6\pm 2\sqrt{19}}{2} \\
& =-3\pm \sqrt{19}
\end{align}\]
Hence, the solution set is\[\left\{ -3\pm \sqrt{19} \right\}\].
