Question 484111


{{{4x^2-7x+3=0}}} Start with the given equation.



Notice that the quadratic {{{4x^2-7x+3}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=4}}}, {{{B=-7}}}, and {{{C=3}}}



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



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



{{{x = (-(-7) +- sqrt( (-7)^2-4(4)(3) ))/(2(4))}}} Plug in  {{{A=4}}}, {{{B=-7}}}, and {{{C=3}}}



{{{x = (7 +- sqrt( (-7)^2-4(4)(3) ))/(2(4))}}} Negate {{{-7}}} to get {{{7}}}. 



{{{x = (7 +- sqrt( 49-4(4)(3) ))/(2(4))}}} Square {{{-7}}} to get {{{49}}}. 



{{{x = (7 +- sqrt( 49-48 ))/(2(4))}}} Multiply {{{4(4)(3)}}} to get {{{48}}}



{{{x = (7 +- sqrt( 1 ))/(2(4))}}} Subtract {{{48}}} from {{{49}}} to get {{{1}}}



{{{x = (7 +- sqrt( 1 ))/(8)}}} Multiply {{{2}}} and {{{4}}} to get {{{8}}}. 



{{{x = (7 +- 1)/(8)}}} Take the square root of {{{1}}} to get {{{1}}}. 



{{{x = (7 + 1)/(8)}}} or {{{x = (7 - 1)/(8)}}} Break up the expression. 



{{{x = (8)/(8)}}} or {{{x =  (6)/(8)}}} Combine like terms. 



{{{x = 1}}} or {{{x = 3/4}}} Simplify. 



So the solutions are {{{x = 1}}} or {{{x = 3/4}}}