Question 159309


{{{3x^2-x-1=0}}} Start with the given equation.



Notice we have a quadratic equation in the form of {{{ax^2+bx+c}}} where {{{a=3}}}, {{{b=-1}}}, 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 = (-(-1) +- sqrt( (-1)^2-4(3)(-1) ))/(2(3))}}} Plug in  {{{a=3}}}, {{{b=-1}}}, and {{{c=-1}}}



{{{x = (1 +- sqrt( (-1)^2-4(3)(-1) ))/(2(3))}}} Negate {{{-1}}} to get {{{1}}}. 



{{{x = (1 +- sqrt( 1-4(3)(-1) ))/(2(3))}}} Square {{{-1}}} to get {{{1}}}. 



{{{x = (1 +- sqrt( 1--12 ))/(2(3))}}} Multiply {{{4(3)(-1)}}} to get {{{-12}}}



{{{x = (1 +- sqrt( 1+12 ))/(2(3))}}} Rewrite {{{sqrt(1--12)}}} as {{{sqrt(1+12)}}}



{{{x = (1 +- sqrt( 13 ))/(2(3))}}} Add {{{1}}} to {{{12}}} to get {{{13}}}



{{{x = (1 +- sqrt( 13 ))/(6)}}} Multiply {{{2}}} and {{{3}}} to get {{{6}}}. 



{{{x = (1+sqrt(13))/(6)}}} or {{{x = (1-sqrt(13))/(6)}}} Break up the expression.  



So the answers are {{{x = (1+sqrt(13))/(6)}}} or {{{x = (1-sqrt(13))/(6)}}} 



which approximate to {{{x=0.768}}} or {{{x=-0.434}}}