Y=2(3b)^2-3

Solution for Y=2(3b)^2-3 equation:

=2(3Y)^2-3

We move all terms to the left:

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

We get rid of parentheses

-23Y^2+3=0

a = -23; b = 0; c = +3;
Δ = b2-4ac
Δ = 02-4·(-23)·3
Δ = 276
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:

$\sqrt{\Delta}=\sqrt{276}=\sqrt{4*69}=\sqrt{4}*\sqrt{69}=2\sqrt{69}$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{69}}{2*-23}=\frac{0-2\sqrt{69}}{-46}
=-\frac{2\sqrt{69}}{-46}
=-\frac{\sqrt{69}}{-23}
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{69}}{2*-23}=\frac{0+2\sqrt{69}}{-46}
=\frac{2\sqrt{69}}{-46}
=\frac{\sqrt{69}}{-23}
$