Question 845307
You could go about it one of two ways:
1) You could multiply both sides of the equation times {{{x^2}}} to get
{{{1+8x+15x^2=0}}} <--> {{{15x^2+8x+1=0}}} and solve that equation.
(That is probably the expected way).
2) You could say {{{y=1/x}}} , and that makes {{{y^2=1/x^2}}} ,
and substituting you would get
{{{y^2+8y+15=0}}} .
Solving that equation you get solutions for {{{y}}} ,
which you can substitute into {{{y=1/x}}} to find the values for {{{x}}}.
 
MULTIPLYING TIMES {{{x^2}}}:
We know that {{{x^2}}} is not zero, because it is in a denominator and if it were zero, {{{1/x^2}}} would not exist. If we found {{{x=0}}} as a solution, we would discard it, because we know it is not a valid solution of
{{{ 1/x^2 + 8/x +15=0}}} .
So we can safely multiply times {{{x^2}}} both sides of the equal sign.will get 
{{{ 1/x^2 + 8/x +15=0}}}
{{{x^2*(1/x^2 + 8/x +15)=x^2*0}}}
{{{ x^2(1/x^2) + x^2*(8/x) +x^2*15=0}}}
{{{1+8x+15x^2=0}}}
{{{15x^2+8x+1=0}}}
The quadratic formula says that the solutions to
{{{ax^2+bx+c=0}}} are given by
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
{{{15x^2+8x+1=0}}} is {{{ax^2+bx+c=0}}} with
{{{a=15}}} , {{{b=8}}} , and {{{c=1}}} .
So {{{x = (-8 +- sqrt(8^2-4*15*1 ))/(2*15) }}}
{{{x = (-8 +- sqrt(64-60))/30 }}}
{{{x = (-8 +- sqrt(4))/30 }}}
{{{x = (-8 +- 2)/30 }}}
The solutions are
{{{x = (-8 + 2)/30 =-6/30=highlight(-1/5)}}} and {{{x = (-8 - 2)/30 =-10/30=highlight(-1/3)}}}
 
NOTE:
You could use the quadratic formula, or you could solve by factoring, or you could even solve by completing the square.
For {{{15x^2+8x+1=0}}} , using the quadratic formula is very reasonable,
because the other options do not seem any easier.
If you had changed variables (option 2) above) to end with
{{{y^2+8y+15=0}}} ,
you may find that factoring is the easiest option.
You may immediately see that you can factor to get
{{{(y+3)(y+5)=0}}} ,
which would tell you that the solutions are {{{y=-3}}} and {{{y=-5}}} .
{{{y=-3}}} means {{{1/x=-3}}} --> {{{1=-3x}}} -->{{{1/(-3)=x}}} --> {{{x=-1/3}}}
{{{y=-5}}} means {{{1/x=-5}}} --> {{{1=-5x}}} -->{{{1/(-5)=x}}} --> {{{x=-1/5}}}