Question 703695
what is the solution of the equation? 
{{{sqrt(2x+13) -5 = x}}}
To solve a radical equation we first need to get the radical, i.e. square root, by itself
Add 5 to both sides to get
{{{sqrt(2x+13)= x + 5}}}
To "undo" the square root function, we need to do the inverse, which is square both sides.  Notice that the square root is now undone
{{{2x+13= (x + 5)^2}}}
Squaring the right side is the same as (x + 5)(x + 5).  Many students often make the mistake of saying it would be {{{x^2 + 25}}} forgetting that they need to FOIL this expression out.  Done correctly we should get
{{{2x+13= x^2 + 10x + 25}}}
We can see this results in a quadratic, which means we have to set it to zero, moving everything to one side
{{{0= x^2 + 10x -2x + 25 - 13}}}
Simplifying the right side by combining like terms yields
{{{0= x^2 + 8x + 12}}}
After set to zero we factor. What multiplies to give 12 but adds to 8?
{{{0= (x + 6)(x + 2)}}}
Set each factor to zero
{{{ x+6 = 0}}} or {{{x + 2=0}}}
and solve for x to get our solutions!
{{{x = -6}}} or {{{x = -2}}}