You didn't attach anything, but this is how you should solve this equation: given an equation [tex]ax^2+bx+c=0[/tex] the solutions are
[tex]x_{1,2} = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
In your case, the values are [tex] a=1,\ b=-8,\ c=7 [/tex]. If you plug these values inside the solving formula, you have
[tex]x_{1,2} = \dfrac{8\pm\sqrt{64-4\cdot 1\cdot 7}}{2\cdot 1} = \dfrac{8\pm\sqrt{64-28}}{2} = \dfrac{8\pm\sqrt{36}}{2} = \dfrac{8\pm6}{2} = 4\pm3[/tex]
So, the two solutions are 1 and 7
