What are the solutions to the equation solvefor x,x^2+2x+y^2+5=0 ?

The solutions to the equation solvefor x,x^2+2x+y^2+5=0 are x=-1+sqrt(-y^2-4),x=-sqrt(-y^2-4)-1

Find the zeros of solvefor x,x^2+2x+y^2+5=0

The zeros of solvefor x,x^2+2x+y^2+5=0 are x=-1+sqrt(-y^2-4),x=-sqrt(-y^2-4)-1

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solve for

x, x^{2} + 2 x + y^{2} + 5 = 0

x = - 1 + - y^{2} - 4, x = - - y^{2} - 4 - 1

Solution steps

x^{2} + 2 x + y^{2} + 5 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 1 \cdot ( y ^{2} + 5 )}{2 \cdot 1}

Simplify

2^{2} - 4 \cdot 1 \cdot (y^{2} + 5) : 2 - y^{2} - 4

x_{1, 2} = \frac{- 2 \pm 2 - y ^{2} - 4}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 2 + 2 - y ^{2} - 4}{2 \cdot 1}, x_{2} = \frac{- 2 - 2 - y ^{2} - 4}{2 \cdot 1}

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

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

The solutions to the quadratic equation are:

x = - 1 + - y^{2} - 4, x = - - y^{2} - 4 - 1

x^{2} + 5 x = 24

5 x^{2} - 2 = 4 x

x (x + 8) = x + 8

x^{2} + (x + 1)^{2} + (x + 2)^{2} = (x + 3)^{2} + (x + 4)^{2}

x^{2} - 12 x + 117 = 0

What are the solutions to the equation solvefor x,x^2+2x+y^2+5=0 ?
The solutions to the equation solvefor x,x^2+2x+y^2+5=0 are x=-1+sqrt(-y^2-4),x=-sqrt(-y^2-4)-1
Find the zeros of solvefor x,x^2+2x+y^2+5=0
The zeros of solvefor x,x^2+2x+y^2+5=0 are x=-1+sqrt(-y^2-4),x=-sqrt(-y^2-4)-1