16y^2-32y-33=0

Solution for 16y^2-32y-33=0 equation:

16y^2-32y-33=0

a = 16; b = -32; c = -33;
Δ = b2-4ac
Δ = -322-4·16·(-33)
Δ = 3136
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3136}=56$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-56}{2*16}=\frac{-24}{32}
=-3/4
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+56}{2*16}=\frac{88}{32}
=2+3/4
$