Question

Please solve: x^2+4x-8=0

Answer 1

$x^2+4x-8=0\\\\\underbrace{x^2+2x\times2+2^2}_{use:(a+b)^2=a^2+2ab+b^2}-2^2-8=0\\\\(x+2)^2-4-8=0\\\\(x+2)^2-12=0\\\\(x+2)^2=12\iff x+2=\sqrt{12}\ or\ x+2=-\sqrt{12}\\\\x=\sqrt{4\times3}-2\ or\ x=-\sqrt{4\times3}-2\\\\x=\sqrt4\times\sqrt3-2\ or\ x=-\sqrt4\times\sqrt3-2\\\\\center\boxed{x=2\sqrt3-2\ or\ x=-2\sqrt3-2}$

Answer 2

 x^2+4x-8=0Δ = 16 + 32 = 48 
x1 =( -4 - 4√2)/2 = -2 - 2√2
x2 =(-4 + 4√2)/2 = -2 + 2√2