Question 42227
{{{sqrt(7x+29)=x+3}}}


Squaring both sides
{{{(7x+29)=(x+3)^2}}}
or {{{7x+29=x^2 + 6x + 9}}}
or {{{x^2 + 6x + 9 - 7x - 29 = 0}}}
or {{{x^2 - x - 20 = 0}}}
or {{{x^2 - 5x + 4x - 20 = 0}}}
or {{{x(x - 5) + 4(x - 5) = 0}}}
or {{{(x+4)(x-5)=0}}}


Hence either (x + 4) = 0 i.e. x = -4 or (x - 5) = 0 i.e. x = 5.


So either x = -4 or x = 5.