Y=3t^2-12t+4

Solution for Y=3t^2-12t+4 equation:

=3Y^2-12Y+4

We move all terms to the left:

-(3Y^2-12Y+4)=0

We get rid of parentheses

-3Y^2+12Y-4=0

a = -3; b = 12; c = -4;
Δ = b2-4ac
Δ = 122-4·(-3)·(-4)
Δ = 96
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:

$\sqrt{\Delta}=\sqrt{96}=\sqrt{16*6}=\sqrt{16}*\sqrt{6}=4\sqrt{6}$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-4\sqrt{6}}{2*-3}=\frac{-12-4\sqrt{6}}{-6}
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4\sqrt{6}}{2*-3}=\frac{-12+4\sqrt{6}}{-6}
$