Y=3x2+12x+3

Solution for Y=3x2+12x+3 equation:

=3Y^2+12Y+3

We move all terms to the left:

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

We get rid of parentheses

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

a = -3; b = -12; c = -3;
Δ = b2-4ac
Δ = -122-4·(-3)·(-3)
Δ = 108
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:

$\sqrt{\Delta}=\sqrt{108}=\sqrt{36*3}=\sqrt{36}*\sqrt{3}=6\sqrt{3}$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-6\sqrt{3}}{2*-3}=\frac{12-6\sqrt{3}}{-6}
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+6\sqrt{3}}{2*-3}=\frac{12+6\sqrt{3}}{-6}
$