The quadratic formula states that the solutions to the equation [tex] ax^2+bx+c=0 [/tex] are (if they exist)
[tex] x_{1,2} = \frac{-b\pm\sqrt{b^2-4ac}}{2a} [/tex]
In your case, the quadratic equation is identified by the choice [tex] a = 1, b=-4, c=-8 [/tex], and this leads to the solutions
[tex] x_{1,2} = \frac{4\pm\sqrt{16+32}}{2} = \frac{4\pm\sqrt{48}}{2} [/tex]
Since [tex] 48 = 16\cdot 3[/tex], we have
[tex] \sqrt{48} = \sqrt{16\cdot 3} = \sqrt{16}\codt \sqrt{3} = 4\sqrt{3} [/tex]
So, the expression for the solutions becomes
[tex] \frac{4\pm4\sqrt{3}}{2} = 2\pm2\sqrt{3} [/tex]
So, the two solutions are [tex] 2 + 2\sqrt{3} [/tex] and [tex] 2 - 2\sqrt{3} [/tex]
