a^2-20a+32=0

Solution for a^2-20a+32=0 equation:

a^2-20a+32=0

a = 1; b = -20; c = +32;
Δ = b2-4ac
Δ = -202-4·1·32
Δ = 272
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{272}=\sqrt{16*17}=\sqrt{16}*\sqrt{17}=4\sqrt{17}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-4\sqrt{17}}{2*1}=\frac{20-4\sqrt{17}}{2}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4\sqrt{17}}{2*1}=\frac{20+4\sqrt{17}}{2}
$