2x^2-96x+576=0

Solution for 2x^2-96x+576=0 equation:

2x^2-96x+576=0

a = 2; b = -96; c = +576;
Δ = b2-4ac
Δ = -962-4·2·576
Δ = 4608
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:

$\sqrt{\Delta}=\sqrt{4608}=\sqrt{2304*2}=\sqrt{2304}*\sqrt{2}=48\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-96)-48\sqrt{2}}{2*2}=\frac{96-48\sqrt{2}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-96)+48\sqrt{2}}{2*2}=\frac{96+48\sqrt{2}}{4}
$