y^2-54y+240=0

Solution for y^2-54y+240=0 equation:

y^2-54y+240=0

a = 1; b = -54; c = +240;
Δ = b2-4ac
Δ = -542-4·1·240
Δ = 1956
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:

$\sqrt{\Delta}=\sqrt{1956}=\sqrt{4*489}=\sqrt{4}*\sqrt{489}=2\sqrt{489}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-54)-2\sqrt{489}}{2*1}=\frac{54-2\sqrt{489}}{2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-54)+2\sqrt{489}}{2*1}=\frac{54+2\sqrt{489}}{2}
$