Question 195129
{{{3x^2-16x+5=0}}} Start with the given equation.



Notice we have a quadratic equation in the form of {{{ax^2+bx+c=0}}} where {{{a=3}}}, {{{b=-16}}}, and {{{c=5}}}



Let's use the quadratic formula to solve for x



{{{x = (-b +- sqrt( b^2-4ac ))/(2a)}}} Start with the quadratic formula



{{{x = (-(-16) +- sqrt( (-16)^2-4(3)(5) ))/(2(3))}}} Plug in  {{{a=3}}}, {{{b=-16}}}, and {{{c=5}}}



{{{x = (16 +- sqrt( (-16)^2-4(3)(5) ))/(2(3))}}} Negate {{{-16}}} to get {{{16}}}. 



{{{x = (16 +- sqrt( 256-4(3)(5) ))/(2(3))}}} Square {{{-16}}} to get {{{256}}}. 



{{{x = (16 +- sqrt( 256-60 ))/(2(3))}}} Multiply {{{4(3)(5)}}} to get {{{60}}}



{{{x = (16 +- sqrt( 196 ))/(2(3))}}} Subtract {{{60}}} from {{{256}}} to get {{{196}}}



{{{x = (16 +- sqrt( 196 ))/(6)}}} Multiply {{{2}}} and {{{3}}} to get {{{6}}}. 



{{{x = (16 +- 14)/(6)}}} Take the square root of {{{196}}} to get {{{14}}}. 



{{{x = (16 + 14)/(6)}}} or {{{x = (16 - 14)/(6)}}} Break up the expression. 



{{{x = (30)/(6)}}} or {{{x =  (2)/(6)}}} Combine like terms. 



{{{x = 5}}} or {{{x = 1/3}}} Simplify. 



So the solutions are {{{x = 5}}} or {{{x = 1/3}}}