32y^2-24y-48=0

Solution for 32y^2-24y-48=0 equation:

32y^2-24y-48=0

a = 32; b = -24; c = -48;
Δ = b2-4ac
Δ = -242-4·32·(-48)
Δ = 6720
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:

$\sqrt{\Delta}=\sqrt{6720}=\sqrt{64*105}=\sqrt{64}*\sqrt{105}=8\sqrt{105}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-8\sqrt{105}}{2*32}=\frac{24-8\sqrt{105}}{64}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+8\sqrt{105}}{2*32}=\frac{24+8\sqrt{105}}{64}
$