Can someone help me with this math equation?

you cant do much here


(2 x^2-3 y) (2 x^2+3 y)

(A + B) * (A - B)
= A^2 - AB + AB - B^2
=A^2-B^2

A = 2x^2
B = 3y

by ^4 do you mean to the power of 4? If so then you factorise it as the difference of two squares.
You need to rewrite it as say a to the power of 2 - b to the power of 2.ie a^2 - b^2 which in your case
4x^4 will be (2X^2)^2 and the 9y^2 will be (3y)^2 . Factors then are (a + b)(a - b) so in your case it then becomes (2x^2 + 3y) (2x^2 - 3y) . You can check then by multiplying these factors out and see do you get your original equation.

(2x^2-3y)(2x^2+3y)
God bless you.

4x^4 - 9y^2= (2x^2)^2 - (3y)^2 {difference of two squares} you will find this:
= (2x^2-3y)(2x^2+3y)

NOTE:a^2-b^2=(a-b)(a+b)

factor 4x square minus 9y square
4x^2-9y^2 = (2x-3y)*(2x+3y).
Solving for y would yield two roots: y = 2x/3 and y = -2x/3.

Looks like the difference of two squares.
Do you happen to know a formula for the difference of two squares?