Y=6x^2+12x+3

Solution for Y=6x^2+12x+3 equation:

=6Y^2+12Y+3

We move all terms to the left:

-(6Y^2+12Y+3)=0

We get rid of parentheses

-6Y^2-12Y-3=0

a = -6; b = -12; c = -3;
Δ = b2-4ac
Δ = -122-4·(-6)·(-3)
Δ = 72
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:

$\sqrt{\Delta}=\sqrt{72}=\sqrt{36*2}=\sqrt{36}*\sqrt{2}=6\sqrt{2}$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-6\sqrt{2}}{2*-6}=\frac{12-6\sqrt{2}}{-12}
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+6\sqrt{2}}{2*-6}=\frac{12+6\sqrt{2}}{-12}
$