Question

What are the solutions of this equation? 2x^2 + 16x − 8 = 0

Answer 1

Answer:

$x=-4\pm2\sqrt{5}$

Step-by-step explanation:

$2x^2+16x-8=0\\\\x^2+8x-4=0\\\\x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\ \\x=\frac{-8\pm\sqrt{(8)^2-4(1)(-4)}}{2(1)}\\ \\x=\frac{-8\pm\sqrt{64+16}}{2}\\\\x=\frac{-8\pm\sqrt{80}}{2}\\ \\x=\frac{-8\pm4\sqrt{5}}{2}\\ \\x=-4\pm2\sqrt{5}$