X^2-32x+96=0

Solution for X^2-32x+96=0 equation:

X^2-32X+96=0

a = 1; b = -32; c = +96;
Δ = b2-4ac
Δ = -322-4·1·96
Δ = 640
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:

$\sqrt{\Delta}=\sqrt{640}=\sqrt{64*10}=\sqrt{64}*\sqrt{10}=8\sqrt{10}$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-8\sqrt{10}}{2*1}=\frac{32-8\sqrt{10}}{2}
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+8\sqrt{10}}{2*1}=\frac{32+8\sqrt{10}}{2}
$