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

The solutions to the equation solvefor x,2y-x^2=xy are x=-(y+sqrt(y^2+8y))/2 ,x=-(y-sqrt(y^2+8y))/2

Find the zeros of solvefor x,2y-x^2=xy

The zeros of solvefor x,2y-x^2=xy are x=-(y+sqrt(y^2+8y))/2 ,x=-(y-sqrt(y^2+8y))/2

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solve for

x, 2 y - x^{2} = x y

x = - \frac{y + y ^{2} + 8 y}{2}, x = - \frac{y - y ^{2} + 8 y}{2}

Solution steps

2 y - x^{2} = x y

Move

x y

to the left side

2 y - x^{2} - x y = 0

Write in the standard form

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

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

Solve with the quadratic formula

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

Simplify

(- y)^{2} - 4 (- 1) \cdot 2 y : y^{2} + 8 y

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

Separate the solutions

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

x = \frac{- ( - y ) + y ^{2} + 8 y}{2 ( - 1 )} : - \frac{y + y ^{2} + 8 y}{2}

x = \frac{- ( - y ) - y ^{2} + 8 y}{2 ( - 1 )} : - \frac{y - y ^{2} + 8 y}{2}

The solutions to the quadratic equation are:

x = - \frac{y + y ^{2} + 8 y}{2}, x = - \frac{y - y ^{2} + 8 y}{2}

y, 2 x y + 5 y^{2} = 4

5 t^{2} + 14 t - 3 = 0

100 = (x - 1) (200 - 40 x)

\frac{4 a}{9} a = 5

x^{2} - 6 x - 2 = 0

What are the solutions to the equation solvefor x,2y-x^2=xy ?
The solutions to the equation solvefor x,2y-x^2=xy are x=-(y+sqrt(y^2+8y))/2 ,x=-(y-sqrt(y^2+8y))/2
Find the zeros of solvefor x,2y-x^2=xy
The zeros of solvefor x,2y-x^2=xy are x=-(y+sqrt(y^2+8y))/2 ,x=-(y-sqrt(y^2+8y))/2