Y=3x^2+6x-11

Solution for Y=3x^2+6x-11 equation:

=3Y^2+6Y-11

We move all terms to the left:

-(3Y^2+6Y-11)=0

We get rid of parentheses

-3Y^2-6Y+11=0

a = -3; b = -6; c = +11;
Δ = b2-4ac
Δ = -62-4·(-3)·11
Δ = 168
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:

$\sqrt{\Delta}=\sqrt{168}=\sqrt{4*42}=\sqrt{4}*\sqrt{42}=2\sqrt{42}$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{42}}{2*-3}=\frac{6-2\sqrt{42}}{-6}
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{42}}{2*-3}=\frac{6+2\sqrt{42}}{-6}
$