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

The answer to solvefor y,x=sqrt(y-3)+3 is y=x^2-6x+12 {x>= 3}

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

solve for

y, x = y - 3 + 3

y = x^{2} - 6 x + 12 {x \geq 3}

Solution steps

x = y - 3 + 3

Switch sides

y - 3 + 3 = x

Move

3

to the right side

y - 3 = x - 3

Square both sides:

y - 3 = x^{2} - 6 x + 9

y - 3 = x^{2} - 6 x + 9

Solve

y - 3 = x^{2} - 6 x + 9 : y = x^{2} - 6 x + 12

y = x^{2} - 6 x + 12

Verify Solutions:

y = x^{2} - 6 x + 12 {x \geq 3}

The solution is

y = x^{2} - 6 x + 12 {x \geq 3}

x^{\frac{3}{4}} = \frac{1}{27}

7 x + 8 = - 8

(5 x + 6)^{\frac{1}{4}} = - 2

x = 2 + 8 - x

What is the answer to solvefor y,x=sqrt(y-3)+3 ?
The answer to solvefor y,x=sqrt(y-3)+3 is y=x^2-6x+12 {x>= 3}