2x^2-4x-64=0

Solution for 2x^2-4x-64=0 equation:

2x^2-4x-64=0

a = 2; b = -4; c = -64;
Δ = b2-4ac
Δ = -42-4·2·(-64)
Δ = 528
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{528}=\sqrt{16*33}=\sqrt{16}*\sqrt{33}=4\sqrt{33}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4\sqrt{33}}{2*2}=\frac{4-4\sqrt{33}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{33}}{2*2}=\frac{4+4\sqrt{33}}{4}
$