Question 342746
{{{4x^2-8x+3=0}}} Start with the given equation.



Notice that the quadratic {{{4x^2-8x+3}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=4}}}, {{{B=-8}}}, 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 = (-(-8) +- sqrt( (-8)^2-4(4)(3) ))/(2(4))}}} Plug in  {{{A=4}}}, {{{B=-8}}}, and {{{C=3}}}



{{{x = (8 +- sqrt( (-8)^2-4(4)(3) ))/(2(4))}}} Negate {{{-8}}} to get {{{8}}}. 



{{{x = (8 +- sqrt( 64-4(4)(3) ))/(2(4))}}} Square {{{-8}}} to get {{{64}}}. 



{{{x = (8 +- sqrt( 64-48 ))/(2(4))}}} Multiply {{{4(4)(3)}}} to get {{{48}}}



{{{x = (8 +- sqrt( 16 ))/(2(4))}}} Subtract {{{48}}} from {{{64}}} to get {{{16}}}



{{{x = (8 +- sqrt( 16 ))/(8)}}} Multiply {{{2}}} and {{{4}}} to get {{{8}}}. 



{{{x = (8 +- 4)/(8)}}} Take the square root of {{{16}}} to get {{{4}}}. 



{{{x = (8 + 4)/(8)}}} or {{{x = (8 - 4)/(8)}}} Break up the expression. 



{{{x = (12)/(8)}}} or {{{x =  (4)/(8)}}} Combine like terms. 



{{{x = 3/2}}} or {{{x = 1/2}}} Simplify. 



So the solutions are {{{x = 3/2}}} or {{{x = 1/2}}} 



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