12=3(3y^2+1)

Solution for 12=3(3y^2+1) equation:

12=3(3y^2+1)

We move all terms to the left:

12-(3(3y^2+1))=0


We calculate terms in parentheses: -(3(3y^2+1)), so:

3(3y^2+1)

We multiply parentheses

9y^2+3

Back to the equation:

-(9y^2+3)


We get rid of parentheses

-9y^2-3+12=0

We add all the numbers together, and all the variables

-9y^2+9=0

a = -9; b = 0; c = +9;
Δ = b2-4ac
Δ = 02-4·(-9)·9
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-18}{2*-9}=\frac{-18}{-18}
=1
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+18}{2*-9}=\frac{18}{-18}
=-1
$