2x^2-32x+75=0

Solution for 2x^2-32x+75=0 equation:

2x^2-32x+75=0

a = 2; b = -32; c = +75;
Δ = b2-4ac
Δ = -322-4·2·75
Δ = 424
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:

$\sqrt{\Delta}=\sqrt{424}=\sqrt{4*106}=\sqrt{4}*\sqrt{106}=2\sqrt{106}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-2\sqrt{106}}{2*2}=\frac{32-2\sqrt{106}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+2\sqrt{106}}{2*2}=\frac{32+2\sqrt{106}}{4}
$