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

The answer to solvefor x,f(x-3)=sqrt(x-3)-3 is

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solvefor x,f(x-3)=sqrt(x-3)-3

solve for

x, f (x - 3) = x - 3 - 3

x = \frac{6 f ^{2} - 6 f + 1 + - 12 f + 1}{2 f ^{2}} {0 < f \leq \frac{1}{12}}, x = \frac{6 f ^{2} - 6 f + 1 - - 12 f + 1}{2 f ^{2}} {f < 0 or 0 < f \leq \frac{1}{12}}

Solution steps

f (x - 3) = x - 3 - 3

Expand

f (x - 3) : x f - 3 f

x f - 3 f = x - 3 - 3

Remove square roots

x^{2} f^{2} - 6 x f^{2} + 6 x f + 9 f^{2} - 18 f + 9 = x - 3

Solve

x^{2} f^{2} - 6 x f^{2} + 6 x f + 9 f^{2} - 18 f + 9 = x - 3 : x = \frac{6 f ^{2} - 6 f + 1 + - 12 f + 1}{2 f ^{2}}, x = \frac{6 f ^{2} - 6 f + 1 - - 12 f + 1}{2 f ^{2}}

x = \frac{6 f ^{2} - 6 f + 1 + - 12 f + 1}{2 f ^{2}}, x = \frac{6 f ^{2} - 6 f + 1 - - 12 f + 1}{2 f ^{2}}

Verify Solutions:

x = \frac{6 f ^{2} - 6 f + 1 + - 12 f + 1}{2 f ^{2}} {0 < f \leq \frac{1}{12}}, x = \frac{6 f ^{2} - 6 f + 1 - - 12 f + 1}{2 f ^{2}} {f < 0 or 0 < f \leq \frac{1}{12}}

The solutions are

x = \frac{6 f ^{2} - 6 f + 1 + - 12 f + 1}{2 f ^{2}} {0 < f \leq \frac{1}{12}}, x = \frac{6 f ^{2} - 6 f + 1 - - 12 f + 1}{2 f ^{2}} {f < 0 or 0 < f \leq \frac{1}{12}}

x - 4 x - 7 = 3

x^{2} = 8 x

\frac{2}{x ^{\frac{1}{3}}} - 2 = 0

a, x = a b

2 x + 6 = x + 3

What is the answer to solvefor x,f(x-3)=sqrt(x-3)-3 ?
The answer to solvefor x,f(x-3)=sqrt(x-3)-3 is