Question 309012
Use the method of your choice.
{{{sqrt(7x+29)}}} = x + 3
Square both sides
7x + 29 = (x + 3)^2
FOIL the right sides
7x + 29 = x^2 + 6x + 9
Combine like terms on the right
0 = x^2 + 6x - 7x + 9 - 29
A quadratic equation
x^2 - x - 20 = 0
Factors to
(x-5)(x+4) = 0
two solutions
x = +5
x = -4
;
check both solution in original equation
x = 5
{{{sqrt(7(5)+29)}}} = 5 + 3 
{{{sqrt(35+29)}}} = 5 + 3
{{{sqrt(64)}}} = 8
8 = 8, a good solution
:
x=-4
{{{sqrt(7(-4)+29)}}} = -4 + 3 
{{{sqrt(-28+29)}}} = -4 + 3
{{{sqrt(1)}}} = -1
+1 does not = -1, not a good solution