Y=-16x^2+32x+12

Solution for Y=-16x^2+32x+12 equation:

=-16Y^2+32Y+12

We move all terms to the left:

-(-16Y^2+32Y+12)=0

We get rid of parentheses

16Y^2-32Y-12=0

a = 16; b = -32; c = -12;
Δ = b2-4ac
Δ = -322-4·16·(-12)
Δ = 1792
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:

$\sqrt{\Delta}=\sqrt{1792}=\sqrt{256*7}=\sqrt{256}*\sqrt{7}=16\sqrt{7}$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-16\sqrt{7}}{2*16}=\frac{32-16\sqrt{7}}{32}
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+16\sqrt{7}}{2*16}=\frac{32+16\sqrt{7}}{32}
$