Answer
the solution set is\[\left\{ -2,5 \right\}\].
Work Step by Step
The given equation can be written as \[{{x}^{2}}-3x-10=0\].
Compare the given equation with the equation\[a{{x}^{2}}+bx+c=0\].
Here,\[a=1,\,\,b=-3,\text{ and }c=-10\].
Now put these values in the quadratic formula: \[x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
That is,
\[\begin{align}
& x=\frac{3\pm \sqrt{{{\left( -3 \right)}^{2}}-4\times 1\times \left( -10 \right)}}{2\times 1} \\
& =\frac{3\pm \sqrt{9+40}}{2} \\
& =\frac{3\pm \sqrt{49}}{2} \\
& =\frac{3\pm 7}{2}
\end{align}\]
Further simplify it to get,
\[\begin{align}
& x=\frac{3\pm 7}{2} \\
& =\frac{3-7}{2},\frac{3+7}{2} \\
& =-\frac{4}{2},\frac{10}{2} \\
& =-2,5
\end{align}\]
Hence, the solution set is\[\left\{ -2,5 \right\}\].
