y^2-24y+20=0

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

y^2-24y+20=0

a = 1; b = -24; c = +20;
Δ = b2-4ac
Δ = -242-4·1·20
Δ = 496
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{496}=\sqrt{16*31}=\sqrt{16}*\sqrt{31}=4\sqrt{31}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-4\sqrt{31}}{2*1}=\frac{24-4\sqrt{31}}{2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+4\sqrt{31}}{2*1}=\frac{24+4\sqrt{31}}{2}
$