Question 875278
{{{2/5+x/3=1/x}}}
 
We could say we use a common denominator of {{{5*3*x=15x}}} :
{{{2/5+x/3=1/x}}}
{{{6x/15x+5x^2/15x=15/15x}}}
{{{(6x+5x^2)/15x=15/15x}}}
{{{6x+5x^2=15}}}
 
Or we could say that we multiply times {{{5*3*x=15x}}} :
{{{2/5+x/3=1/x}}}
{{{15x(2/5+x/3)=15x(1/x)}}}
{{{15x(2/5)+15x(x/3)=15x/x}}}
{{{30x/5+15x^2/3=15}}}
{{{6x+5x^2=15}}}

 
It's the same thing, we end up with
{{{6x+5x^2=15}}} <---> {{{6x+5x^2-15=0}}}  <---> {{{5x^2+6x-15=0}}} anyway,
and now we have to solve that quadratic equation.
Factoring does not work, so we have two options to solve that equation:
 
USING THE QUADRATIC FORMULA:
The quadratic formula says that the solutions (real or imaginary) of
{{{ax^2+bx+c=0}}} are given by
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} .
In the case of {{{5x^2+6x-15=0}}} ,
{{{a=5}}} , {{{b=6}}} , and {{{c=-15}}} , so
{{{x=(-6 +- sqrt(6^2-4*5*(-15) ))/(2*5)=(-6 +- sqrt(36+300))/10=(-6 +- sqrt(336))/10}}}
That can be simplified as
{{{x=(-6 +- sqrt(16*21))/10=(-6 +- sqrt(16)*sqrt(21))/10=(-6 +- 4sqrt(21))/10=(-3 +- 2sqrt(21))/5}}}
 
COMPLETING THE SQUARE:
{{{5x^2+6x-15=0}}}
{{{5x^2+6x=15}}}
{{{(5x^2+6x)/5=15/5}}}
{{{x^2+(6/5)x=3}}}
{{{x^2+2(3/5)x+(3/5)^2=3+(3/5)^2}}}
{{{(x+3/5)^2=75/25+9/25}}}
{{{(x+3/5)^2=84/25}}}
Either {{{x+3/5=sqrt(84/25)=sqrt(4*21)/sqrt(25)=sqrt(4)sqrt(21)/5=2sqrt(21)/5}}}
or {{{x+3/5=-2sqrt(21)/5}}}
{{{x+3/5=2sqrt(21)/5}}}-->{{{x=-3/5+2sqrt(21)/5=(-3+2sqrt(21))5}}}
{{{x+3/5=-2sqrt(21)/5}}}-->{{{x=-3/5-2sqrt(21)/5=(-3-2sqrt(21))5}}}
and both solutions can be expressed as {{{x=(-3 +- 2sqrt(21))/5}}}