What are the solutions to the equation solvefor y,x2+y2+4x-2y+5=0 ?

The solutions to the equation solvefor y,x2+y2+4x-2y+5=0 are y=1+sqrt(-x^2-4x-4),y=1-sqrt(-x^2-4x-4)

Find the zeros of solvefor y,x2+y2+4x-2y+5=0

The zeros of solvefor y,x2+y2+4x-2y+5=0 are y=1+sqrt(-x^2-4x-4),y=1-sqrt(-x^2-4x-4)

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solve for

y, x^{2} + y^{2} + 4 x - 2 y + 5 = 0

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

Solution steps

x^{2} + y^{2} + 4 x - 2 y + 5 = 0

Write in the standard form

a x^{2} + b x + c = 0

y^{2} - 2 y + x^{2} + 4 x + 5 = 0

Solve with the quadratic formula

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

Simplify

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

15 X - 10 = 6 X - (X + 2) + (- X + 3)

7^{- 2} = x

2 - v = 5 (v + 4)

3 (x + 2) = 3 + 2 (x - 1)

5 \frac{y}{9} - \frac{y}{9} = \frac{8}{9}

What are the solutions to the equation solvefor y,x2+y2+4x-2y+5=0 ?
The solutions to the equation solvefor y,x2+y2+4x-2y+5=0 are y=1+sqrt(-x^2-4x-4),y=1-sqrt(-x^2-4x-4)
Find the zeros of solvefor y,x2+y2+4x-2y+5=0
The zeros of solvefor y,x2+y2+4x-2y+5=0 are y=1+sqrt(-x^2-4x-4),y=1-sqrt(-x^2-4x-4)