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

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

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

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

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solve for

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

y = 1 + x + 1, y = 1 - x + 1

Solution steps

x = y^{2} - 2 y

Switch sides

y^{2} - 2 y = x

Move

x

to the left side

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

Solve with the quadratic formula

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

Simplify

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

y = 1 + x + 1, y = 1 - x + 1

2 a^{2} + a = 5

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

20 = 45 t - 5 t^{2}

6 x^{2} + 23 x + 7 = 0

2 x^{2} = 3 x - 5

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