Y+3=3x^2+27x

Solution for Y+3=3x^2+27x equation:

+3=3Y^2+27Y

We move all terms to the left:

+3-(3Y^2+27Y)=0

We get rid of parentheses

-3Y^2-27Y+3=0

a = -3; b = -27; c = +3;
Δ = b2-4ac
Δ = -272-4·(-3)·3
Δ = 765
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:

$\sqrt{\Delta}=\sqrt{765}=\sqrt{9*85}=\sqrt{9}*\sqrt{85}=3\sqrt{85}$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-27)-3\sqrt{85}}{2*-3}=\frac{27-3\sqrt{85}}{-6}
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-27)+3\sqrt{85}}{2*-3}=\frac{27+3\sqrt{85}}{-6}
$