Answer
the solution set is\[\left\{ \frac{1\pm \sqrt{57}}{2} \right\}\].
Work Step by Step
The given equation can be written as\[{{x}^{2}}-x-14=0\].
Compare the given equation with the equation\[a{{x}^{2}}+bx+c=0\].
Here,\[a=1,\,\,b=-1,\text{ and }c=-14\]
Now put these values in the quadratic formula: \[x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
That is,
\[\begin{align}
& x=\frac{1\pm \sqrt{{{\left( -1 \right)}^{2}}-4\times 1\times \left( -14 \right)}}{2\times 1} \\
& =\frac{1\pm \sqrt{1+56}}{2} \\
& =\frac{1\pm \sqrt{57}}{2}
\end{align}\]
Hence, the solution set is\[\left\{ \frac{1\pm \sqrt{57}}{2} \right\}\].
