Question

Solve the following quadratic equation:
2x^2+4x–8=0

Answer 1

Answer:

it does not have roots it can solve by shreedherr acharya formula

I hope it may help u

Answer 2

$2x^2+4x-8=0\\2(x^2+2x-4)=0\\x^2+2x-4=0$

$D=\sqrt{b^2-4ac}\\ D=\sqrt{4-4*1*(-4)}\\ D=\sqrt{4+16}\\ D=\sqrt{20}\\ D=2\sqrt{5}$

$x_1=\frac{-b+\sqrt{D} }{2a}\\\\x_1=\frac{-2+(2\sqrt{5} )}{2}\\ x_1=-1+\sqrt{5}$

$x_2=\frac{-b-\sqrt{D} }{2a}\\ \\x_2=\frac{-2-2\sqrt{5} }{2} \\\\x_2=-1-\sqrt{5}$

Here, x₁ and x₂ represents the two zeroes of the equation.

Hope it helps...  :D