-y^2+48y-480=0

Solution for -y^2+48y-480=0 equation:

-y^2+48y-480=0

We add all the numbers together, and all the variables

-1y^2+48y-480=0

a = -1; b = 48; c = -480;
Δ = b2-4ac
Δ = 482-4·(-1)·(-480)
Δ = 384
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:

$\sqrt{\Delta}=\sqrt{384}=\sqrt{64*6}=\sqrt{64}*\sqrt{6}=8\sqrt{6}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-8\sqrt{6}}{2*-1}=\frac{-48-8\sqrt{6}}{-2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+8\sqrt{6}}{2*-1}=\frac{-48+8\sqrt{6}}{-2}
$