r^2-32r+12=0

Solution for r^2-32r+12=0 equation:

r^2-32r+12=0

a = 1; b = -32; c = +12;
Δ = b2-4ac
Δ = -322-4·1·12
Δ = 976
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{976}=\sqrt{16*61}=\sqrt{16}*\sqrt{61}=4\sqrt{61}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-4\sqrt{61}}{2*1}=\frac{32-4\sqrt{61}}{2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+4\sqrt{61}}{2*1}=\frac{32+4\sqrt{61}}{2}
$