Okay, here we have this:
[tex]x^2+2x-7=0[/tex]We will solve using the general formula, then we obtain:
[tex]\begin{gathered} x_{1,2}=\frac{-2\pm\sqrt[]{2^2-4\cdot1\cdot(-7)}}{2\cdot1} \\ =\frac{-2\pm\sqrt[]{4+28}}{2} \\ =\frac{-2\pm\sqrt[]{32}}{2} \\ =\frac{-2\pm4\sqrt[]{2}}{2} \end{gathered}[/tex]Let's separate the solutions:
[tex]\begin{gathered} x_1=\frac{-2+4\sqrt[]{2}}{2} \\ =\frac{2(-1+2\sqrt[]{2})}{2} \\ =-1+2\sqrt[]{2} \end{gathered}[/tex][tex]\begin{gathered} x_2_{}=\frac{-2-4\sqrt[]{2}}{2} \\ =\frac{2(-1-2\sqrt[]{2})}{2} \\ =-1-2\sqrt[]{2} \end{gathered}[/tex]Finally we obtain that the roots are: -1+2√2 and -1-2√2.
