Question

Решите уравнение 
x^2+4x-8=0

Answer 1

$x^{2} +4x-8=0 \ D=16+32=48 \ sqrt{D} = sqrt{48} = sqrt{16*3} =4 sqrt{3} \ x_{1} = frac{-4-4 sqrt{3} }{2} = frac{2(-2-2 sqrt{3}) }{2} =-2-2 sqrt{3} \ x_{2} = frac{-4+4 sqrt{3} }{2} = frac{2(-2+2 sqrt{3}) }{2} =-2+2 sqrt{3}$