y^2+48y+288=0

Solution for y^2+48y+288=0 equation:

y^2+48y+288=0

a = 1; b = 48; c = +288;
Δ = b2-4ac
Δ = 482-4·1·288
Δ = 1152
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:

$\sqrt{\Delta}=\sqrt{1152}=\sqrt{576*2}=\sqrt{576}*\sqrt{2}=24\sqrt{2}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-24\sqrt{2}}{2*1}=\frac{-48-24\sqrt{2}}{2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+24\sqrt{2}}{2*1}=\frac{-48+24\sqrt{2}}{2}
$