Question 171807


{{{3x^2-x-2=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=-2}}}



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)(-2) ))/(2(3))}}} Plug in  {{{a=3}}}, {{{b=-1}}}, and {{{c=-2}}}



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



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



{{{x = (1 +- sqrt( 1--24 ))/(2(3))}}} Multiply {{{4(3)(-2)}}} to get {{{-24}}}



{{{x = (1 +- sqrt( 1+24 ))/(2(3))}}} Rewrite {{{sqrt(1--24)}}} as {{{sqrt(1+24)}}}



{{{x = (1 +- sqrt( 25 ))/(2(3))}}} Add {{{1}}} to {{{24}}} to get {{{25}}}



{{{x = (1 +- sqrt( 25 ))/(6)}}} Multiply {{{2}}} and {{{3}}} to get {{{6}}}. 



{{{x = (1 +- 5)/(6)}}} Take the square root of {{{25}}} to get {{{5}}}. 



{{{x = (1 + 5)/(6)}}} or {{{x = (1 - 5)/(6)}}} Break up the expression. 



{{{x = (6)/(6)}}} or {{{x =  (-4)/(6)}}} Combine like terms. 



{{{x = 1}}} or {{{x = -2/3}}} Simplify. 



So the answers are {{{x = 1}}} or {{{x = -2/3}}}