What is the answer to solvefor x,4xy-3y^2+12y+4=0 ?

The answer to solvefor x,4xy-3y^2+12y+4=0 is x=(3y^2-12y-4)/(4y);\quad y\ne 0

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solvefor x,4xy-3y^2+12y+4=0

solve for

x, 4 x y - 3 y^{2} + 12 y + 4 = 0

x = \frac{3 y ^{2} - 12 y - 4}{4 y}; y \neq = 0

Solution steps

4 x y - 3 y^{2} + 12 y + 4 = 0

Move

3 y^{2}

to the right side

4 x y + 12 y + 4 = 3 y^{2}

Move

12 y

to the right side

4 x y + 4 = 3 y^{2} - 12 y

Move

4

to the right side

4 x y = 3 y^{2} - 12 y - 4

Divide both sides by

4 y; y \neq = 0

x = \frac{3 y ^{2} - 12 y - 4}{4 y}; y \neq = 0

x, 2 x = 12 + 2 y

x, 3 x + 2 y + 8 = 0

x, - 2 x + 3 y = 3

x, - 2 x + 3 y = 2

3 x + 5 = 2 x

What is the answer to solvefor x,4xy-3y^2+12y+4=0 ?
The answer to solvefor x,4xy-3y^2+12y+4=0 is x=(3y^2-12y-4)/(4y);\quad y\ne 0