2y(y-3)=27+y2

Solution for 2y(y-3)=27+y2 equation:

2y(y-3)=27+y2

We move all terms to the left:

2y(y-3)-(27+y2)=0

We add all the numbers together, and all the variables

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

We multiply parentheses

-(+y^2+27)+2y^2-6y=0

We get rid of parentheses

-y^2+2y^2-6y-27=0

We add all the numbers together, and all the variables

y^2-6y-27=0

a = 1; b = -6; c = -27;
Δ = b2-4ac
Δ = -62-4·1·(-27)
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-12}{2*1}=\frac{-6}{2}
=-3
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+12}{2*1}=\frac{18}{2}
=9
$