Question 210149
I'm assuming you meant to write {{{3x^2-5x+4 = 0}}}





{{{3x^2-5x+4=0}}} Start with the given equation.



Notice that the quadratic {{{3x^2-5x+4}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=3}}}, {{{B=-5}}}, and {{{C=4}}}



Let's use the quadratic formula to solve for "x":



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



{{{x = (-(-5) +- sqrt( (-5)^2-4(3)(4) ))/(2(3))}}} Plug in  {{{A=3}}}, {{{B=-5}}}, and {{{C=4}}}



{{{x = (5 +- sqrt( (-5)^2-4(3)(4) ))/(2(3))}}} Negate {{{-5}}} to get {{{5}}}. 



{{{x = (5 +- sqrt( 25-4(3)(4) ))/(2(3))}}} Square {{{-5}}} to get {{{25}}}. 



{{{x = (5 +- sqrt( 25-48 ))/(2(3))}}} Multiply {{{4(3)(4)}}} to get {{{48}}}



{{{x = (5 +- sqrt( -23 ))/(2(3))}}} Subtract {{{48}}} from {{{25}}} to get {{{-23}}}



{{{x = (5 +- sqrt( -23 ))/(6)}}} Multiply {{{2}}} and {{{3}}} to get {{{6}}}. 



{{{x = (5 +- i*sqrt(23))/(6)}}} Simplify the square root  



{{{x = (5+i*sqrt(23))/(6)}}} or {{{x = (5-i*sqrt(23))/(6)}}} Break up the expression.  



So the solutions are {{{x = (5+i*sqrt(23))/(6)}}} or {{{x = (5-i*sqrt(23))/(6)}}}