Y=-16x^2+12x+3

Solution for Y=-16x^2+12x+3 equation:

=-16Y^2+12Y+3

We move all terms to the left:

-(-16Y^2+12Y+3)=0

We get rid of parentheses

16Y^2-12Y-3=0

a = 16; b = -12; c = -3;
Δ = b2-4ac
Δ = -122-4·16·(-3)
Δ = 336
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:

$\sqrt{\Delta}=\sqrt{336}=\sqrt{16*21}=\sqrt{16}*\sqrt{21}=4\sqrt{21}$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4\sqrt{21}}{2*16}=\frac{12-4\sqrt{21}}{32}
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4\sqrt{21}}{2*16}=\frac{12+4\sqrt{21}}{32}
$