What is 3x²-[7x-(4x-x²)=3]?

Question

Is y²-1/9= (y X 1/3) +(y X 1/3)
Also what is 2x²+x-6 I think I am suppose to simply it

Answer 1

First, I believe you miswrote the problem because there is an equal sign inside of parentheses.
This is probably right:

Find x if 3x²-[7x-(4x-x²)]=3?

3x²-[7x-(4x-x²)]=3
3x²-[7x-(4x-x²)] -3 = 0
3x²-7x+(4x-x²) -3 = 0
3x²-7x+4x-x² -3 = 0
2x²-3x -3 = 0
x² - 3/2x -3/2 = 0
x² - 3/2x +9/4 -9/4 -3/2 = 0
(x² - 3/2x +9/16) -9/16 -3/2 = 0
(x - 3/4)² -9/16 -3/2 = 0
(x - 3/4)² -33/16 = 0
(x - 3/4)² = 33/16
x - 3/4 = +/-(33)^(1/2)/4
x = 3/4 +/-(33)^(1/2)/4
x = (1/4)(3 +/- (33)^(1/2))
so,
x = (1/4)(3 + (33)^(1/2)), or, x = (1/4)(3 - (33)^(1/2))