For this you can use the quadratic formula, which is used to solve any equation of the second degree:
[tex]\begin{gathered} x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}\Rightarrow\text{ Quadratic formula} \\ \text{ For a quadratic equation in standard form } \\ ax^2+bx+c=0 \end{gathered}[/tex]So, in this case, you have
[tex]\begin{gathered} x^2+2x-8=0 \\ a=1 \\ b=2 \\ c=-8 \end{gathered}[/tex][tex]\begin{gathered} x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ x=\frac{-2\pm\sqrt[]{(2)^2-4(1)(-8)}}{2(1)} \\ x=\frac{-2\pm\sqrt[]{4+32}}{2} \\ x=\frac{-2\pm\sqrt[]{36}}{2} \\ x=\frac{-2\pm6}{2} \end{gathered}[/tex]Then, the solutions will be
[tex]\begin{gathered} x_1=\frac{-2+6}{2} \\ x_1=\frac{4}{2} \\ x_1=2 \end{gathered}[/tex]and
[tex]\begin{gathered} x_2=\frac{-2-6}{2} \\ x_2=\frac{-8}{2} \\ x_2=-4 \end{gathered}[/tex]Therefore, the solutions of the quadratic equation will be
[tex]\begin{gathered} x_1=2 \\ \text{ and} \\ x_2=-4 \end{gathered}[/tex]