```Question 249025

{{{sqrt(2x+5)=3+sqrt(x-2)}}} Add {{{sqrt(x-2)}}} to both sides.

{{{(sqrt(2x+5))^2=(3+sqrt(x-2))^2}}} Square both sides.

{{{2x+5=(3+sqrt(x-2))^2}}} Square the square root on the left side to eliminate it.

{{{2x+5=(3+sqrt(x-2))(3+sqrt(x-2))}}} Expand

{{{2x+5=3*3+3*sqrt(x-2)+3*sqrt(x-2)+sqrt(x-2)*sqrt(x-2)}}} FOIL

{{{2x+5=9+3*sqrt(x-2)+3*sqrt(x-2)+(sqrt(x-2))^2}}} Multiply

{{{2x+5=9+3*sqrt(x-2)+3*sqrt(x-2)+x-2}}}  Square the last square root on the right side to eliminate it.

{{{2x+5=7+x+6*sqrt(x-2)}}} Combine like terms.

{{{2x+5-7-x=6*sqrt(x-2)}}} Move all non square root terms to one side.

{{{x-2=6*sqrt(x-2)}}} Combine like terms.

{{{(x-2)^2=(6*sqrt(x-2))^2}}} Square both sides.

{{{(x-2)^2=36(x-2)}}} Square {{{6*sqrt(x-2)}}} to get {{{36(x-2)}}}

{{{x^2-4x+4=36(x-2)}}} FOIL

{{{x^2-4x+4=36x-72}}} Distribute

{{{x^2-4x+4-36x+72=0}}} Get every term to the left side.

{{{x^2-40x+76=0}}} Combine like terms.

Notice that the quadratic {{{x^2-40x+76}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=1}}}, {{{B=-40}}}, and {{{C=76}}}

Let's use the quadratic formula to solve for "x":

{{{x = (-(-40) +- sqrt( (-40)^2-4(1)(76) ))/(2(1))}}} Plug in  {{{A=1}}}, {{{B=-40}}}, and {{{C=76}}}

{{{x = (40 +- sqrt( (-40)^2-4(1)(76) ))/(2(1))}}} Negate {{{-40}}} to get {{{40}}}.

{{{x = (40 +- sqrt( 1600-4(1)(76) ))/(2(1))}}} Square {{{-40}}} to get {{{1600}}}.

{{{x = (40 +- sqrt( 1600-304 ))/(2(1))}}} Multiply {{{4(1)(76)}}} to get {{{304}}}

{{{x = (40 +- sqrt( 1296 ))/(2(1))}}} Subtract {{{304}}} from {{{1600}}} to get {{{1296}}}

{{{x = (40 +- sqrt( 1296 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}.

{{{x = (40 +- 36)/(2)}}} Take the square root of {{{1296}}} to get {{{36}}}.

{{{x = (40 + 36)/(2)}}} or {{{x = (40 - 36)/(2)}}} Break up the expression.

{{{x = (76)/(2)}}} or {{{x =  (4)/(2)}}} Combine like terms.

{{{x = 38}}} or {{{x = 2}}} Simplify.

So the solutions are {{{x = 38}}} or {{{x = 2}}}```