Question 1153074
 {{{x=2sqrt(2x+5)}}}........first square both sides to get rid of sqrt

{{{x^2=2^2*(2x+5)}}}

{{{x^2=4(2x+5)}}}

{{{x^2=8x+20}}}

{{{x^2-8x-20=0}}}.....factor completely: write {{{-8x}}} as {{{-2x+10x}}}

{{{x^2-2x+10x-20=0}}}.......group

{{{(x^2-2x)+(10x-20)=0}}}

{{{x(x-2)+10(x-2)=0}}}

{{{(x - 10)(x + 2) = 0}}}

solutions:

if {{{(x - 10) = 0}}}=>{{{x=10}}}

if {{{(x + 2) = 0}}}=>{{{x=-2}}}


since you need solution that will make {{{x=2sqrt(2x+5)}}} true, check the solutions by plugging them into {{{x=2sqrt(2x+5)}}}


{{{x=2sqrt(2x+5)}}}=>{{{x=10}}}

{{{10=2sqrt(2*10+5)}}}

{{{10=2sqrt(25)}}}

{{{10=2*5}}}

{{{10=10}}}=>{{{true}}}


{{{x=2sqrt(2x+5)}}}=>{{{x=-2}}}

{{{-2=2sqrt(2*(-2)+5)}}}

{{{-2=2sqrt(-4+5)}}}

{{{-2=2sqrt(1)}}}

{{{-2=2*1}}}

{{{-2=2}}}=>{{{false}}}


=> solution to {{{x=2sqrt(2x+5)}}} is {{{x=10}}}