2x^2-64x+252=0

Solution for 2x^2-64x+252=0 equation:

2x^2-64x+252=0

a = 2; b = -64; c = +252;
Δ = b2-4ac
Δ = -642-4·2·252
Δ = 2080
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:

$\sqrt{\Delta}=\sqrt{2080}=\sqrt{16*130}=\sqrt{16}*\sqrt{130}=4\sqrt{130}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-64)-4\sqrt{130}}{2*2}=\frac{64-4\sqrt{130}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-64)+4\sqrt{130}}{2*2}=\frac{64+4\sqrt{130}}{4}
$