Can someone help with this elimination question?

6x^2 - 3x = 2y - 5 and 2x^2 + x = y - 4. Double the 2nd equation:
4x^2 + 2x = 2y - 8, and subtract that from the 1st equation:
2x^2 - 5x = 3. Collecting terms on 1 side, you have:
2x^2 - 5x - 3 = 0, or
(2x + 1)(x - 3) = 0, so
x = -1/2 or x = 3.
Then substitute each of these back in, to get a corresponding value of y for each solution.

6x^2 - 3x = 2y - 5
2x^2 + x = y - 4.
4x^2 + 2x = 2y - 8
6x^2 - 3x + 5 = 4x^2 + 2x + 8
2X^2 - 5x - 3 = 0
2x^2 - 6x + x - 3 = 0
2x(x - 3) + (x - 3) = 0
(2x + 1)(x - 3) = 0
Solutions:
x = -1/2, y = 4
x = 3, y = 25

multiply the 2nd equation by -2 and add the two equations together...

(-4x^2-2x=-2y+8)+(6x^2-3x=2y-5) which equals

2x^2-5x=3

2x^2-5x-3=0

2x^2+x-6x-3=0

x(2x+1)-3(2x+1)=0

(x-3)(2x+1)=0 so x=3 and -1/2

using 6x^2-3x=2y-5 with the xs found above we find the corresponding ys

x=3, y=25 (3, 25)

x=-.5, y=4 (-0.5, 4)