x^2-32x+160=0

Solution for x^2-32x+160=0 equation:

x^2-32x+160=0

a = 1; b = -32; c = +160;
Δ = b2-4ac
Δ = -322-4·1·160
Δ = 384
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:

$\sqrt{\Delta}=\sqrt{384}=\sqrt{64*6}=\sqrt{64}*\sqrt{6}=8\sqrt{6}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-8\sqrt{6}}{2*1}=\frac{32-8\sqrt{6}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+8\sqrt{6}}{2*1}=\frac{32+8\sqrt{6}}{2}
$