Question 471848
{{{(x-3)(x^2+3x-2)=0}}}.

we know if {{{x-3=0}}}...->.....{{{x=3}}}........first solution


take {{{x^2+3x-2=0}}} and complete the square


{{{x^2+3x=2}}}.....to the both sides add  square of {{{1/2}}} of coefficient of middle term (which is {{{3}}}): {{{((1/2)3)^2=(3/2)^2}}}


{{{x^2+3x(3/2)^2=2+(3/2)^2}}}

{{{(x+3/2)^2=2+9/4}}}

{{{(x+3/2)^2=17/4}}}

now we can apply square root method to find other two solutions

if {{{u^2=d}}}, then {{{u=sqrt(d)}}}  or {{{u=-sqrt(d)}}}

in your case {{{u=(x+3/2)}}} and {{{d=17/4}}}

consequently

{{{x+3/2=sqrt(17/4)}}}  

{{{x+1.5=sqrt(4.25)}}} 

{{{x=2.06-1.5}}} 

{{{x=0.56}}} ......second solution


or 

{{{x+3/2=-sqrt(17/4)}}}  

{{{x+1.5=-sqrt(4.25)}}} 

{{{x=-2.06-1.5}}} 

{{{x=-3.56}}} ........third solution

check:

{{{(x-3)(x^2+3x-2)=0}}}...{{{x=3}}}........first solution

{{{(3-3)(3^2+3*3-2)=0}}}

{{{(0)(9+9-2)=0}}}

{{{0=0}}}


{{{(x-3)(x^2+3x-2)=0}}}......{{{x=0.56}}} ......second solution


{{{(0.56-3)((0.56)^2+3(0.56)-2)=0}}}


{{{(-2.44)(0.3136+1.68-2)=0}}}

{{{(-2.44)(1.9936-2)=0}}}

{{{(-2.44)(2-2)=0}}}

{{{(-2.44)(0)=0}}}

{{{0=0}}}

you can check third  one by yourself