2y^2+24y+48=0

Solution for 2y^2+24y+48=0 equation:

2y^2+24y+48=0

a = 2; b = 24; c = +48;
Δ = b2-4ac
Δ = 242-4·2·48
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*2}=\frac{-24-8\sqrt{3}}{4}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+8\sqrt{3}}{2*2}=\frac{-24+8\sqrt{3}}{4}
$