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

The solutions to the equation solvefor x,3y^2-2x^2=2xy are x=-((1+sqrt(7))y)/2 ,x=-((1-sqrt(7))y)/2

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

The zeros of solvefor x,3y^2-2x^2=2xy are x=-((1+sqrt(7))y)/2 ,x=-((1-sqrt(7))y)/2

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solve for

x, 3 y^{2} - 2 x^{2} = 2 x y

x = - \frac{( 1 + 7 ) y}{2}, x = - \frac{( 1 - 7 ) y}{2}

Solution steps

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

Move

2 x y

to the left side

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Simplify

(- 2 y)^{2} - 4 (- 2) \cdot 3 y^{2} : 27 y

x_{1, 2} = \frac{- ( - 2 y ) \pm 2 7 y}{2 ( - 2 )}

Separate the solutions

x_{1} = \frac{- ( - 2 y ) + 2 7 y}{2 ( - 2 )}, x_{2} = \frac{- ( - 2 y ) - 2 7 y}{2 ( - 2 )}

x = \frac{- ( - 2 y ) + 2 7 y}{2 ( - 2 )} : - \frac{( 1 + 7 ) y}{2}

x = \frac{- ( - 2 y ) - 2 7 y}{2 ( - 2 )} : - \frac{( 1 - 7 ) y}{2}

The solutions to the quadratic equation are:

x = - \frac{( 1 + 7 ) y}{2}, x = - \frac{( 1 - 7 ) y}{2}

10 x^{2} - x + 11 = 0

y^{2} + 4 y + 6 = 0

2 (x - 3)^{2} = 18

3 g^{2} + 4 g - 9 = 0

x, z = - 4 x^{2} - 6 x y - 3 y^{2} + 2 x + y + 10

What are the solutions to the equation solvefor x,3y^2-2x^2=2xy ?
The solutions to the equation solvefor x,3y^2-2x^2=2xy are x=-((1+sqrt(7))y)/2 ,x=-((1-sqrt(7))y)/2
Find the zeros of solvefor x,3y^2-2x^2=2xy
The zeros of solvefor x,3y^2-2x^2=2xy are x=-((1+sqrt(7))y)/2 ,x=-((1-sqrt(7))y)/2