solve for y: 2y-3=√3y^2-10y+12?

2y - 3 = √(3y² - 10y + 12)
[2y - 3]² = [√(3y² - 10y + 12)]²
4y² - 12y + 9 = 3y² - 10y + 12
y² - 2y - 3 = 0
(y - 3)(y + 1) = 0
y = 3 or y = -1 (rejected)
Answer : y = 3


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Check :

When y = 3 :
L.H.S. = 2*3 - 3 = 3
R.H.S. = √(3*3² - 10*3 + 12) = 3
L.H.S. = R.H.S.
Hence, y = 3 is accepted.

When y = -1 :
L.H.S. = 2*(-1) - 3 = -5
R.H.S. = √[3*(-1)² - 10*(-1) + 12] = 5
L.H.S. ≠ R.H.S.
Hence, y = -1 is rejected.

3y^2-10y+12 = 4y^2 - 12y+9 , with y>3/2