Question 104057
Transforming the equation,

{{{x^2-5x-2=0}}}

Note that,perfect squares have the format,

{{{(a+b)^2=a^2+2ab+b^2}}}


Thus, comparing this to the equation,

2ab=-5x--------------------------------(1)

and also,

a=x------------------------------------(2)

Substituting (2) to (1),

2bx=-5x
 2b=-5
b=-5/2

Then,

{{{(a+b)^2=a^2+2ab+b^2=x^2-5x+(25/4)}}}

Thus,


{{{x^2-5x-2=0}}}

{{{x^2-5x+(25/4)=0+2+(25/4)}}}

{{{(x^2+(25/4))^2=33/4}}}

{{{x+(5/2)=+(sqrt(33)/2)}}} or {{{x+(5/2)=-(sqrt(33)/2)}}}
{{{x=-(5/2)+(sqrt(33)/2)}}} or {{{x=-(5/2)-(sqrt(33)/2)}}}