Question 197267
{{{5/(2x-3)=(2x)/(x+4)}}} Start with the given equation.



{{{5(x+4)=2x(2x-3)}}} Cross multiply



{{{5x+20=4x^2-6x}}} Distribute



{{{0=4x^2-6x-5x-20}}} Get all terms to the right side
 


{{{0=4x^2-11x-20}}} Combine like terms.



Notice we have a quadratic in the form of {{{Ax^2+Bx+C}}} where {{{A=4}}}, {{{B=-11}}}, and {{{C=-20}}}



Let's use the quadratic formula to solve for x



{{{x = (-B +- sqrt( B^2-4AC ))/(2A)}}} Start with the quadratic formula



{{{x = (-(-11) +- sqrt( (-11)^2-4(4)(-20) ))/(2(4))}}} Plug in  {{{A=4}}}, {{{B=-11}}}, and {{{C=-20}}}



{{{x = (11 +- sqrt( (-11)^2-4(4)(-20) ))/(2(4))}}} Negate {{{-11}}} to get {{{11}}}. 



{{{x = (11 +- sqrt( 121-4(4)(-20) ))/(2(4))}}} Square {{{-11}}} to get {{{121}}}. 



{{{x = (11 +- sqrt( 121--320 ))/(2(4))}}} Multiply {{{4(4)(-20)}}} to get {{{-320}}}



{{{x = (11 +- sqrt( 121+320 ))/(2(4))}}} Rewrite {{{sqrt(121--320)}}} as {{{sqrt(121+320)}}}



{{{x = (11 +- sqrt( 441 ))/(2(4))}}} Add {{{121}}} to {{{320}}} to get {{{441}}}



{{{x = (11 +- sqrt( 441 ))/(8)}}} Multiply {{{2}}} and {{{4}}} to get {{{8}}}. 



{{{x = (11 +- 21)/(8)}}} Take the square root of {{{441}}} to get {{{21}}}. 



{{{x = (11 + 21)/(8)}}} or {{{x = (11 - 21)/(8)}}} Break up the expression. 



{{{x = (32)/(8)}}} or {{{x =  (-10)/(8)}}} Combine like terms. 



{{{x = 4}}} or {{{x = -5/4}}} Simplify. 



So the solutions are {{{x = 4}}} or {{{x = -5/4}}}