Y=4x^2-3^3

Solution for Y=4x^2-3^3 equation:

=4Y^2-3^3

We move all terms to the left:

-(4Y^2-3^3)=0

We get rid of parentheses

-4Y^2+3^3=0

We add all the numbers together, and all the variables

-4Y^2+27=0

a = -4; b = 0; c = +27;
Δ = b2-4ac
Δ = 02-4·(-4)·27
Δ = 432
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:

$\sqrt{\Delta}=\sqrt{432}=\sqrt{144*3}=\sqrt{144}*\sqrt{3}=12\sqrt{3}$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12\sqrt{3}}{2*-4}=\frac{0-12\sqrt{3}}{-8}
=-\frac{12\sqrt{3}}{-8}
=-\frac{3\sqrt{3}}{-2}
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12\sqrt{3}}{2*-4}=\frac{0+12\sqrt{3}}{-8}
=\frac{12\sqrt{3}}{-8}
=\frac{3\sqrt{3}}{-2}
$