What are the solutions to the equation solvefor y,x^2+2xy-4+y^2=0 ?

The solutions to the equation solvefor y,x^2+2xy-4+y^2=0 are y=-x+2,y=-x-2

Find the zeros of solvefor y,x^2+2xy-4+y^2=0

The zeros of solvefor y,x^2+2xy-4+y^2=0 are y=-x+2,y=-x-2

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solve for

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

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

Solution steps

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

Write in the standard form

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

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

10 x^{2} + 15 x = 0

26 x^{2} + 25 x - 975 = 0

r^{2} + 15 r + 50 = 6 - r

(p + 1)^{2} = \frac{4}{9}

- 16 t^{2} + 100 = 0

What are the solutions to the equation solvefor y,x^2+2xy-4+y^2=0 ?
The solutions to the equation solvefor y,x^2+2xy-4+y^2=0 are y=-x+2,y=-x-2
Find the zeros of solvefor y,x^2+2xy-4+y^2=0
The zeros of solvefor y,x^2+2xy-4+y^2=0 are y=-x+2,y=-x-2