Question 200195


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



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



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



{{{x = (4 +- sqrt( 16-4(4)(3) ))/(2(4))}}} Square {{{-4}}} to get {{{16}}}. 



{{{x = (4 +- sqrt( 16-48 ))/(2(4))}}} Multiply {{{4(4)(3)}}} to get {{{48}}}



{{{x = (4 +- sqrt( -32 ))/(2(4))}}} Subtract {{{48}}} from {{{16}}} to get {{{-32}}}



{{{x = (4 +- sqrt( -32 ))/(8)}}} Multiply {{{2}}} and {{{4}}} to get {{{8}}}. 



{{{x = (4 +- 4i*sqrt(2))/(8)}}} Simplify the square root  (note: If you need help with simplifying square roots, check out this <a href=http://www.algebra.com/algebra/homework/Radicals/simplifying-square-roots.solver> solver</a>)  



{{{x = (4+4i*sqrt(2))/(8)}}} or {{{x = (4-4i*sqrt(2))/(8)}}} Break up the expression. 



{{{x = (1+i*sqrt(2))/(2)}}} or {{{x = (1-i*sqrt(2))/(2)}}} Reduce



So the solutions are {{{x = (1+i*sqrt(2))/(2)}}} or {{{x = (1-i*sqrt(2))/(2)}}}