What are the solutions to the equation solvefor x,-3xy^2-2x^2y+8xy=0 ?

The solutions to the equation solvefor x,-3xy^2-2x^2y+8xy=0 are x=0,x=(-3y+8)/2 ;\quad y\ne 0

Find the zeros of solvefor x,-3xy^2-2x^2y+8xy=0

The zeros of solvefor x,-3xy^2-2x^2y+8xy=0 are x=0,x=(-3y+8)/2 ;\quad y\ne 0

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solve for

x, - 3 x y^{2} - 2 x^{2} y + 8 x y = 0

x = 0, x = \frac{- 3 y + 8}{2}; y \neq = 0

Solution steps

- 3 x y^{2} - 2 x^{2} y + 8 x y = 0

Write in the standard form

a x^{2} + b x + c = 0

- 2 y x^{2} + (- 3 y^{2} + 8 y) x = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 3 y ^{2} + 8 y ) \pm ( - 3 y ^{2} + 8 y ) ^{2} - 4 ( - 2 y ) \cdot 0}{2 ( - 2 y )}

Simplify

(- 3 y^{2} + 8 y)^{2} - 4 (- 2 y) \cdot 0 : - 3 y^{2} + 8 y

x_{1, 2} = \frac{- ( - 3 y ^{2} + 8 y ) \pm ( - 3 y ^{2} + 8 y )}{2 ( - 2 y )}; y \neq = 0

Separate the solutions

x_{1} = \frac{- ( - 3 y ^{2} + 8 y ) - 3 y ^{2} + 8 y}{2 ( - 2 y )}, x_{2} = \frac{- ( - 3 y ^{2} + 8 y ) - ( - 3 y ^{2} + 8 y )}{2 ( - 2 y )}

x = \frac{- ( - 3 y ^{2} + 8 y ) - 3 y ^{2} + 8 y}{2 ( - 2 y )} : 0

x = \frac{- ( - 3 y ^{2} + 8 y ) - ( - 3 y ^{2} + 8 y )}{2 ( - 2 y )} : \frac{- 3 y + 8}{2}

The solutions to the quadratic equation are:

x = 0, x = \frac{- 3 y + 8}{2}; y \neq = 0

- 4.9 x^{2} - 5 x + 450 = 0

7 x^{2} - 3 = 109

5 x^{2} - 8 x = 6

- x^{2} + 15 x + 100 = 0

24 x^{2} + 2 x - 15 = 0

What are the solutions to the equation solvefor x,-3xy^2-2x^2y+8xy=0 ?
The solutions to the equation solvefor x,-3xy^2-2x^2y+8xy=0 are x=0,x=(-3y+8)/2 ;\quad y\ne 0
Find the zeros of solvefor x,-3xy^2-2x^2y+8xy=0
The zeros of solvefor x,-3xy^2-2x^2y+8xy=0 are x=0,x=(-3y+8)/2 ;\quad y\ne 0