Question 337711
1 plus the sqrt of 2x+1 = 3x-8
{{{1+sqrt(2x+1) = 3x -8 }}}
Subtracting 1 from both sides:
{{{sqrt(2x+1) = 3x -9 }}}
Square both sides:
{{{ 2x+1 = (3x-9)^2 }}}
{{{ 2x+1 = (3x-9)(3x-9) }}}
{{{ 2x+1 = 9x^2-27x-27x+81 }}}
{{{ 2x+1 = 9x^2-54x+81 }}}
{{{ 1 = 9x^2-56x+81 }}}
{{{ 0 = 9x^2-56x+80 }}}
Applying quadratic formula we get:
x = {4, 2.22}
Plug both back in to check for "extraneous" solutions.  Doing so, we eliminate the 2.22 leaving us with:
x = 4
.
Details of quadratic formula:
*[invoke quadratic "x", 9, -56, 80 ]