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

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

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

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

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solve for

y, x = y^{2} - z^{2}

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

Solution steps

x = y^{2} - z^{2}

Switch sides

y^{2} - z^{2} = x

Move

z^{2}

to the right side

y^{2} = x + z^{2}

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

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

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

(x + 7)^{2} - 5 (x + 7) - 14 = 0

x^{2} - 128 = 0

(38 - x)^{2} + (19 - x)^{2} = x^{2}

x^{2} + 10 x + 20 = 0

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