What is the answer to solvefor y,x=2sqrt(3-y) ?

The answer to solvefor y,x=2sqrt(3-y) is y=-(x^2-12)/4 {x>= 0}

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solvefor y,x=2sqrt(3-y)

solve for

y, x = 2 3 - y

y = - \frac{x ^{2} - 12}{4} {x \geq 0}

Solution steps

x = 2 3 - y

Switch sides

2 3 - y = x

Divide both sides by

2

3 - y = \frac{x}{2}

Square both sides:

3 - y = \frac{x ^{2}}{4}

3 - y = \frac{x ^{2}}{4}

Solve

3 - y = \frac{x ^{2}}{4} : y = - \frac{x ^{2} - 12}{4}

y = - \frac{x ^{2} - 12}{4}

Verify Solutions:

y = - \frac{x ^{2} - 12}{4} {x \geq 0}

The solution is

y = - \frac{x ^{2} - 12}{4} {x \geq 0}

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

1 - x = 19

y^{2} - 2 y + 1 + y = 5

4096 x = 64 x^{2}

21 - x = x - 1

What is the answer to solvefor y,x=2sqrt(3-y) ?
The answer to solvefor y,x=2sqrt(3-y) is y=-(x^2-12)/4 {x>= 0}