16y^2+64y-48=0

Solution for 16y^2+64y-48=0 equation:

16y^2+64y-48=0

a = 16; b = 64; c = -48;
Δ = b2-4ac
Δ = 642-4·16·(-48)
Δ = 7168
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:

$\sqrt{\Delta}=\sqrt{7168}=\sqrt{1024*7}=\sqrt{1024}*\sqrt{7}=32\sqrt{7}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(64)-32\sqrt{7}}{2*16}=\frac{-64-32\sqrt{7}}{32}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(64)+32\sqrt{7}}{2*16}=\frac{-64+32\sqrt{7}}{32}
$