Question 186687
{{{1-13/x+36/x^2=0}}} Start with the given equation.



{{{x^2-13x+36=0}}} Multiply EVERY term by the LCD {{{x^2}}} to clear out the fractions.



Notice we have a quadratic equation in the form of {{{ax^2+bx+c}}} where {{{a=1}}}, {{{b=-13}}}, and {{{c=36}}}



Let's use the quadratic formula to solve for x



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



{{{x = (-(-13) +- sqrt( (-13)^2-4(1)(36) ))/(2(1))}}} Plug in  {{{a=1}}}, {{{b=-13}}}, and {{{c=36}}}



{{{x = (13 +- sqrt( (-13)^2-4(1)(36) ))/(2(1))}}} Negate {{{-13}}} to get {{{13}}}. 



{{{x = (13 +- sqrt( 169-4(1)(36) ))/(2(1))}}} Square {{{-13}}} to get {{{169}}}. 



{{{x = (13 +- sqrt( 169-144 ))/(2(1))}}} Multiply {{{4(1)(36)}}} to get {{{144}}}



{{{x = (13 +- sqrt( 25 ))/(2(1))}}} Subtract {{{144}}} from {{{169}}} to get {{{25}}}



{{{x = (13 +- sqrt( 25 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{x = (13 +- 5)/(2)}}} Take the square root of {{{25}}} to get {{{5}}}. 



{{{x = (13 + 5)/(2)}}} or {{{x = (13 - 5)/(2)}}} Break up the expression. 



{{{x = (18)/(2)}}} or {{{x =  (8)/(2)}}} Combine like terms. 



{{{x = 9}}} or {{{x = 4}}} Simplify. 



So the solutions are {{{x = 9}}} or {{{x = 4}}}