Y=3x^2+9x-4

Solution for Y=3x^2+9x-4 equation:

=3Y^2+9Y-4

We move all terms to the left:

-(3Y^2+9Y-4)=0

We get rid of parentheses

-3Y^2-9Y+4=0

a = -3; b = -9; c = +4;
Δ = b2-4ac
Δ = -92-4·(-3)·4
Δ = 129
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-\sqrt{129}}{2*-3}=\frac{9-\sqrt{129}}{-6}
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+\sqrt{129}}{2*-3}=\frac{9+\sqrt{129}}{-6}
$