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

The solutions to the equation solvefor y,x=y^2-y are y=(1+sqrt(1+4x))/2 ,y=(1-sqrt(1+4x))/2

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

The zeros of solvefor y,x=y^2-y are y=(1+sqrt(1+4x))/2 ,y=(1-sqrt(1+4x))/2

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solve for

y, x = y^{2} - y

y = \frac{1 + 1 + 4 x}{2}, y = \frac{1 - 1 + 4 x}{2}

Solution steps

x = y^{2} - y

Switch sides

y^{2} - y = x

Move

x

to the left side

y^{2} - y - x = 0

Solve with the quadratic formula

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

Simplify

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

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

Separate the solutions

y_{1} = \frac{- ( - 1 ) + 1 + 4 x}{2 \cdot 1}, y_{2} = \frac{- ( - 1 ) - 1 + 4 x}{2 \cdot 1}

y = \frac{- ( - 1 ) + 1 + 4 x}{2 \cdot 1} : \frac{1 + 1 + 4 x}{2}

y = \frac{- ( - 1 ) - 1 + 4 x}{2 \cdot 1} : \frac{1 - 1 + 4 x}{2}

The solutions to the quadratic equation are:

y = \frac{1 + 1 + 4 x}{2}, y = \frac{1 - 1 + 4 x}{2}

1 - 3 x^{2} = 2 (x - 1)

m (x) = 3 x^{2} + x - 5, m (\frac{1}{2})

- x^{2} = - 81 x

(11 x - 1)^{2} - 2 (11 x - 1) - 3 = 0

26 t^{2} - 355 t + 300 = 0

What are the solutions to the equation solvefor y,x=y^2-y ?
The solutions to the equation solvefor y,x=y^2-y are y=(1+sqrt(1+4x))/2 ,y=(1-sqrt(1+4x))/2
Find the zeros of solvefor y,x=y^2-y
The zeros of solvefor y,x=y^2-y are y=(1+sqrt(1+4x))/2 ,y=(1-sqrt(1+4x))/2