Question 165055
sqrt{7x + 29} = x+3 
There is nothing to move over as the radical symbol is isolated already. 
We square both sides: 
(sqrt{7x + 29})^2 = (x+3)^2 
7x + 29 = (x+3) (x+3) 
7x + 29 = x^2 + 6x + 9 
Bring the entire left side to the right side and set entire equation to = 0. 
x^2 + 6x + 9 - 7x - 29 = 0 
We now combine like terms. 
x^2 - x - 20 = 0 
We factor this quadratic equation. 
(x + 4) (x - 5) = 0 
Set each factor to = 0 and solve for x. 
x + 4 = 0 
x = -4 
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x - 5 = 0 
x = 5 
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We have two answers for x: x = -4 and x = 5 but are they both acceptable 

answers? In other words, if we replace x in the original equation with -4 and 
5, will the answer be the same on both sides of the equation? 
The only way to know is by CHECKING. 

CHECKING:

Let x = - 4.

sqrt{7x + 29} = x + 3....origiginal question

sqrt{7(-4) + 29} = (-4) + 3

sqrt{-28 + 29} = -1

sqrt{1} = -1

1 DOES NOT EQUAL -1.

We REJECT x = -4 because it is does not produce the same answer on both sides of the equation after checking.

Now, let x = 5 and do the same thing.

sqrt{7x + 29} = x + 3....original question

sqrt{7(5) + 29} = 5 + 3

sqrt{35 + 29} = 8

sqrt{64} = 8

8 = 8...It checks!!

The only TRUE answer here is x = 8