What are the solutions to the equation solvefor y,x=y^2+2y-1 ?

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

Find the zeros of solvefor y,x=y^2+2y-1

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

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solve for

y, x = y^{2} + 2 y - 1

y = - 1 + x + 2, y = - x + 2 - 1

Solution steps

x = y^{2} + 2 y - 1

Switch sides

y^{2} + 2 y - 1 = x

Move

x

to the left side

y^{2} + 2 y - 1 - x = 0

Solve with the quadratic formula

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

Simplify

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

y = - 1 + x + 2, y = - x + 2 - 1

x^{2} - 2 x + 6 = 2 x + 11

x^{2} = 8 x - 15

x^{2} - 6 x - 71 = 0

4 c^{2} + 17 c + 12 = 0

- 3 x^{2} - 3 x + 36 = 0

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