<PRE><B>Question 150473</B>
{{{sqrt(7x+29)}}} = x + 3
:
Square both sides, this gets rid of the radical on the left
7x + 29 = (x+3)^2
:
FOIL (x+3)(x+3)
7x + 29 = x^2 + 6x + 9
:
0 = x^2 + 6x - 7x + 9 - 29
:
A quadratic equation:
x^2 - x - 20 = 0
:
Factors to:
(x-5)(x+4) = 0
:
The positive solution 
x = 5
:
Check in the original equation
{{{sqrt(7(5)+29)}}} = 5 + 3
{{{sqrt(64)}}} = 8
:
The negative solution
x = -4
Check in original equation
{{{sqrt(7(-4)+29)}}} = -4 + 3
{{{sqrt(-28+29)}}} = -1
{{{sqrt(1)}}} = -1
+1 does not = -1
:
Only x=5 is a solution </PRE>