Question 186688
{{{1+3/x=5/x^2}}} Start with the given equation. 



{{{x^2+3x=5}}} Multiply EVERY term by the LCD {{{x^2}}} to clear out the fractions.



{{{x^2+3x-5=0}}} Subtract 5 from both sides.



Notice we have a quadratic equation in the form of {{{ax^2+bx+c}}} where {{{a=1}}}, {{{b=3}}}, and {{{c=-5}}}



Let's use the quadratic formula to solve for x



{{{x = (-b +- sqrt( b^2-4ac ))/(2a)}}} Start with the quadratic formula



{{{x = (-(3) +- sqrt( (3)^2-4(1)(-5) ))/(2(1))}}} Plug in  {{{a=1}}}, {{{b=3}}}, and {{{c=-5}}}



{{{x = (-3 +- sqrt( 9-4(1)(-5) ))/(2(1))}}} Square {{{3}}} to get {{{9}}}. 



{{{x = (-3 +- sqrt( 9--20 ))/(2(1))}}} Multiply {{{4(1)(-5)}}} to get {{{-20}}}



{{{x = (-3 +- sqrt( 9+20 ))/(2(1))}}} Rewrite {{{sqrt(9--20)}}} as {{{sqrt(9+20)}}}



{{{x = (-3 +- sqrt( 29 ))/(2(1))}}} Add {{{9}}} to {{{20}}} to get {{{29}}}



{{{x = (-3 +- sqrt( 29 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{x = (-3+sqrt(29))/(2)}}} or {{{x = (-3-sqrt(29))/(2)}}} Break up the expression.  



So the answers are {{{x = (-3+sqrt(29))/(2)}}} or {{{x = (-3-sqrt(29))/(2)}}} 



which approximate to {{{x=1.193}}} or {{{x=-4.193}}}