Y=(3z^2)-16z+16

Solution for Y=(3z^2)-16z+16 equation:

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

We move all terms to the left:

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

We get rid of parentheses

-3Y^2+16Y-16=0

a = -3; b = 16; c = -16;
Δ = b2-4ac
Δ = 162-4·(-3)·(-16)
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-8}{2*-3}=\frac{-24}{-6}
=+4
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+8}{2*-3}=\frac{-8}{-6}
=1+1/3
$