Question 172191


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



Notice we have a quadratic equation 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 answers are {{{x = 3/2}}} or {{{x = 1/2}}}