What are the solutions to the equation solvefor x=y^2+4y,y ?

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

Find the zeros of solvefor x=y^2+4y,y

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

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solve for

x = y^{2} + 4 y, y

y = - 2 + x + 4, y = - x + 4 - 2

Solution steps

x = y^{2} + 4 y

Switch sides

y^{2} + 4 y = x

Move

x

to the left side

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

Solve with the quadratic formula

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

Simplify

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

y = - 2 + x + 4, y = - x + 4 - 2

- x^{2} + 5 x - 2

- x^{2} + 5 x + 3

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

2 x^{2} - 3 x - 16 = 16

2 x^{2} = 4 x^{2} - 32

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