32y^2-16y-48=0

Solution for 32y^2-16y-48=0 equation:

32y^2-16y-48=0

a = 32; b = -16; c = -48;
Δ = b2-4ac
Δ = -162-4·32·(-48)
Δ = 6400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{6400}=80$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-80}{2*32}=\frac{-64}{64}
=-1
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+80}{2*32}=\frac{96}{64}
=1+1/2
$