2y=3y^2-1

Solution for 2y=3y^2-1 equation:

2y=3y^2-1

We move all terms to the left:

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

We get rid of parentheses

-3y^2+2y+1=0

a = -3; b = 2; c = +1;
Δ = b2-4ac
Δ = 22-4·(-3)·1
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-4}{2*-3}=\frac{-6}{-6}
=1
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+4}{2*-3}=\frac{2}{-6}
=-1/3
$