Y=3x^2-18x+16

Solution for Y=3x^2-18x+16 equation:

=3Y^2-18Y+16

We move all terms to the left:

-(3Y^2-18Y+16)=0

We get rid of parentheses

-3Y^2+18Y-16=0

a = -3; b = 18; c = -16;
Δ = b2-4ac
Δ = 182-4·(-3)·(-16)
Δ = 132
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:

$\sqrt{\Delta}=\sqrt{132}=\sqrt{4*33}=\sqrt{4}*\sqrt{33}=2\sqrt{33}$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-2\sqrt{33}}{2*-3}=\frac{-18-2\sqrt{33}}{-6}
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+2\sqrt{33}}{2*-3}=\frac{-18+2\sqrt{33}}{-6}
$