-4y^2+16y+12=0

Solution for -4y^2+16y+12=0 equation:

-4y^2+16y+12=0

a = -4; b = 16; c = +12;
Δ = b2-4ac
Δ = 162-4·(-4)·12
Δ = 448
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:

$\sqrt{\Delta}=\sqrt{448}=\sqrt{64*7}=\sqrt{64}*\sqrt{7}=8\sqrt{7}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-8\sqrt{7}}{2*-4}=\frac{-16-8\sqrt{7}}{-8}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+8\sqrt{7}}{2*-4}=\frac{-16+8\sqrt{7}}{-8}
$