Question 294259


{{{6x^2-9x+1=0}}} Start with the given equation.



Notice that the quadratic {{{6x^2-9x+1}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=6}}}, {{{B=-9}}}, and {{{C=1}}}



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



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



{{{x = (-(-9) +- sqrt( (-9)^2-4(6)(1) ))/(2(6))}}} Plug in  {{{A=6}}}, {{{B=-9}}}, and {{{C=1}}}



{{{x = (9 +- sqrt( (-9)^2-4(6)(1) ))/(2(6))}}} Negate {{{-9}}} to get {{{9}}}. 



{{{x = (9 +- sqrt( 81-4(6)(1) ))/(2(6))}}} Square {{{-9}}} to get {{{81}}}. 



{{{x = (9 +- sqrt( 81-24 ))/(2(6))}}} Multiply {{{4(6)(1)}}} to get {{{24}}}



{{{x = (9 +- sqrt( 57 ))/(2(6))}}} Subtract {{{24}}} from {{{81}}} to get {{{57}}}



{{{x = (9 +- sqrt( 57 ))/(12)}}} Multiply {{{2}}} and {{{6}}} to get {{{12}}}. 



{{{x = (9+sqrt(57))/(12)}}} or {{{x = (9-sqrt(57))/(12)}}} Break up the expression.  



So the solutions are {{{x = (9+sqrt(57))/(12)}}} or {{{x = (9-sqrt(57))/(12)}}}