x^2-32x+100=0

Solution for x^2-32x+100=0 equation:

x^2-32x+100=0

a = 1; b = -32; c = +100;
Δ = b2-4ac
Δ = -322-4·1·100
Δ = 624
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:

$\sqrt{\Delta}=\sqrt{624}=\sqrt{16*39}=\sqrt{16}*\sqrt{39}=4\sqrt{39}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-4\sqrt{39}}{2*1}=\frac{32-4\sqrt{39}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+4\sqrt{39}}{2*1}=\frac{32+4\sqrt{39}}{2}
$