Answer
$\left\{ -2-\sqrt{11},-2+\sqrt{11} \right\}$
Work Step by Step
The standard form of the given equation, $
(x+5)(x-1)=2
,$ is
\begin{array}{l}\require{cancel}
x(x)+x(-1)+5(x)+5(-1)=2
\\\\
x^2-x+5x-5=2
\\\\
x^2+(-x+5x)+(-5-2)=0
\\\\
x^2+4x-7=0
.\end{array}
Using $\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ (or the Quadratic Formula), the solutions of the quadratic equation, $
x^2+4x-7=0
,$ are
\begin{array}{l}\require{cancel}
\dfrac{-(4)\pm\sqrt{(4)^2-4(1)(-7)}}{2(1)}
\\\\=
\dfrac{-4\pm\sqrt{16+28}}{2}
\\\\=
\dfrac{-4\pm\sqrt{44}}{2}
\\\\=
\dfrac{-4\pm\sqrt{4\cdot11}}{2}
\\\\=
\dfrac{-4\pm\sqrt{(2)^2\cdot11}}{2}
\\\\=
\dfrac{-4\pm2\sqrt{11}}{2}
\\\\=
\dfrac{2(-2\pm\sqrt{11})}{2}
\\\\=
\dfrac{\cancel{2}(-2\pm\sqrt{11})}{\cancel{2}}
\\\\=
-2\pm\sqrt{11}
.\end{array}
Hence, the solutions are $
\left\{ -2-\sqrt{11},-2+\sqrt{11} \right\}
.$
