Y=-16x^2+12x+4

Solution for Y=-16x^2+12x+4 equation:

=-16Y^2+12Y+4

We move all terms to the left:

-(-16Y^2+12Y+4)=0

We get rid of parentheses

16Y^2-12Y-4=0

a = 16; b = -12; c = -4;
Δ = b2-4ac
Δ = -122-4·16·(-4)
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-20}{2*16}=\frac{-8}{32}
=-1/4
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+20}{2*16}=\frac{32}{32}
=1
$