y^2-24y+96=0

Solution for y^2-24y+96=0 equation:

y^2-24y+96=0

a = 1; b = -24; c = +96;
Δ = b2-4ac
Δ = -242-4·1·96
Δ = 192
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:

$\sqrt{\Delta}=\sqrt{192}=\sqrt{64*3}=\sqrt{64}*\sqrt{3}=8\sqrt{3}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-8\sqrt{3}}{2*1}=\frac{24-8\sqrt{3}}{2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+8\sqrt{3}}{2*1}=\frac{24+8\sqrt{3}}{2}
$