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

The solutions to the equation solvefor y,3y^2-2x^3=xy are y=(x+xsqrt(1+24x))/6 ,y=(x-xsqrt(1+24x))/6

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

The zeros of solvefor y,3y^2-2x^3=xy are y=(x+xsqrt(1+24x))/6 ,y=(x-xsqrt(1+24x))/6

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solve for

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

y = \frac{x + x 1 + 24 x}{6}, y = \frac{x - x 1 + 24 x}{6}

Solution steps

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

Move

x y

to the left side

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Simplify

(- x)^{2} - 4 \cdot 3 (- 2 x^{3}) : x 1 + 24 x

y_{1, 2} = \frac{- ( - x ) \pm x 1 + 24 x}{2 \cdot 3}

Separate the solutions

y_{1} = \frac{- ( - x ) + x 1 + 24 x}{2 \cdot 3}, y_{2} = \frac{- ( - x ) - x 1 + 24 x}{2 \cdot 3}

y = \frac{- ( - x ) + x 1 + 24 x}{2 \cdot 3} : \frac{x + x 1 + 24 x}{6}

y = \frac{- ( - x ) - x 1 + 24 x}{2 \cdot 3} : \frac{x - x 1 + 24 x}{6}

The solutions to the quadratic equation are:

y = \frac{x + x 1 + 24 x}{6}, y = \frac{x - x 1 + 24 x}{6}

64 - x^{2} = - 17

x (x - 11) + 30 = 0

2 n^{2} - 19 n + 17 = - 3 n^{2}

9 x^{2} + 18 x - 68 = 7

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

What are the solutions to the equation solvefor y,3y^2-2x^3=xy ?
The solutions to the equation solvefor y,3y^2-2x^3=xy are y=(x+xsqrt(1+24x))/6 ,y=(x-xsqrt(1+24x))/6
Find the zeros of solvefor y,3y^2-2x^3=xy
The zeros of solvefor y,3y^2-2x^3=xy are y=(x+xsqrt(1+24x))/6 ,y=(x-xsqrt(1+24x))/6