Question 174716


{{{9x^2+40=36x}}} Start with the given equation.



{{{9x^2-36x+40=0}}} Subtract 36x from both sides.



Notice we have a quadratic equation in the form of {{{ax^2+bx+c}}} where {{{a=9}}}, {{{b=-36}}}, and {{{c=40}}}



Let's use the quadratic formula to solve for x



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



{{{x = (-(-36) +- sqrt( (-36)^2-4(9)(40) ))/(2(9))}}} Plug in  {{{a=9}}}, {{{b=-36}}}, and {{{c=40}}}



{{{x = (36 +- sqrt( (-36)^2-4(9)(40) ))/(2(9))}}} Negate {{{-36}}} to get {{{36}}}. 



{{{x = (36 +- sqrt( 1296-4(9)(40) ))/(2(9))}}} Square {{{-36}}} to get {{{1296}}}. 



{{{x = (36 +- sqrt( 1296-1440 ))/(2(9))}}} Multiply {{{4(9)(40)}}} to get {{{1440}}}



{{{x = (36 +- sqrt( -144 ))/(2(9))}}} Subtract {{{1440}}} from {{{1296}}} to get {{{-144}}}



{{{x = (36 +- sqrt( -144 ))/(18)}}} Multiply {{{2}}} and {{{9}}} to get {{{18}}}. 



{{{x = (36 +- 12*i)/(18)}}} Take the square root of {{{-144}}} to get {{{12*i}}}. 



{{{x = (36 + 12*i)/(18)}}} or {{{x = (36 - 12*i)/(18)}}} Break up the expression. 



{{{x = (36)/(18) + (12*i)/(18)}}} or {{{x =  (36)/(18) - (12*i)/(18)}}} Break up the fraction for each case. 



{{{x = 2+(2/3)*i}}} or {{{x = 2-(2/3)*i}}} Reduce. 



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