2x^2+40x+64=0

Solution for 2x^2+40x+64=0 equation:

2x^2+40x+64=0

a = 2; b = 40; c = +64;
Δ = b2-4ac
Δ = 402-4·2·64
Δ = 1088
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:

$\sqrt{\Delta}=\sqrt{1088}=\sqrt{64*17}=\sqrt{64}*\sqrt{17}=8\sqrt{17}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-8\sqrt{17}}{2*2}=\frac{-40-8\sqrt{17}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+8\sqrt{17}}{2*2}=\frac{-40+8\sqrt{17}}{4}
$