What is the answer to solvefor x,f(x-5)=sqrt(x+5)+7 ?

The answer to solvefor x,f(x-5)=sqrt(x+5)+7 is

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solvefor x,f(x-5)=sqrt(x+5)+7

solve for

x, f (x - 5) = x + 5 + 7

x = \frac{10 f ^{2} + 14 f + 1 + 40 f ^{2} + 28 f + 1}{2 f ^{2}} {f > 0}, x = \frac{10 f ^{2} + 14 f + 1 - 40 f ^{2} + 28 f + 1}{2 f ^{2}} {f \leq - \frac{7}{10}}

Solution steps

f (x - 5) = x + 5 + 7

Expand

f (x - 5) : x f - 5 f

x f - 5 f = x + 5 + 7

Remove square roots

x^{2} f^{2} - 10 x f^{2} - 14 x f + 25 f^{2} + 70 f + 49 = x + 5

Solve

x^{2} f^{2} - 10 x f^{2} - 14 x f + 25 f^{2} + 70 f + 49 = x + 5 : x = \frac{10 f ^{2} + 14 f + 1 + 40 f ^{2} + 28 f + 1}{2 f ^{2}}, x = \frac{10 f ^{2} + 14 f + 1 - 40 f ^{2} + 28 f + 1}{2 f ^{2}}

x = \frac{10 f ^{2} + 14 f + 1 + 40 f ^{2} + 28 f + 1}{2 f ^{2}}, x = \frac{10 f ^{2} + 14 f + 1 - 40 f ^{2} + 28 f + 1}{2 f ^{2}}

Verify Solutions:

x = \frac{10 f ^{2} + 14 f + 1 + 40 f ^{2} + 28 f + 1}{2 f ^{2}} {f > 0}, x = \frac{10 f ^{2} + 14 f + 1 - 40 f ^{2} + 28 f + 1}{2 f ^{2}} {f \leq - \frac{7}{10}}

The solutions are

x = \frac{10 f ^{2} + 14 f + 1 + 40 f ^{2} + 28 f + 1}{2 f ^{2}} {f > 0}, x = \frac{10 f ^{2} + 14 f + 1 - 40 f ^{2} + 28 f + 1}{2 f ^{2}} {f \leq - \frac{7}{10}}

5 + 5 (3 x + 4) = 55

x - 2 = x - 8

- 3 = 2 x - 1 + 1 + x

a, a = b

22 - 2 x = 2 + x + 1

What is the answer to solvefor x,f(x-5)=sqrt(x+5)+7 ?
The answer to solvefor x,f(x-5)=sqrt(x+5)+7 is