r^2-20r+24=0

Solution for r^2-20r+24=0 equation:

r^2-20r+24=0

a = 1; b = -20; c = +24;
Δ = b2-4ac
Δ = -202-4·1·24
Δ = 304
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:

$\sqrt{\Delta}=\sqrt{304}=\sqrt{16*19}=\sqrt{16}*\sqrt{19}=4\sqrt{19}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-4\sqrt{19}}{2*1}=\frac{20-4\sqrt{19}}{2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4\sqrt{19}}{2*1}=\frac{20+4\sqrt{19}}{2}
$