Question 318067


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



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



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



{{{x = (-1 +- sqrt( 1--48 ))/(2(6))}}} Multiply {{{4(6)(-2)}}} to get {{{-48}}}



{{{x = (-1 +- sqrt( 1+48 ))/(2(6))}}} Rewrite {{{sqrt(1--48)}}} as {{{sqrt(1+48)}}}



{{{x = (-1 +- sqrt( 49 ))/(2(6))}}} Add {{{1}}} to {{{48}}} to get {{{49}}}



{{{x = (-1 +- sqrt( 49 ))/(12)}}} Multiply {{{2}}} and {{{6}}} to get {{{12}}}. 



{{{x = (-1 +- 7)/(12)}}} Take the square root of {{{49}}} to get {{{7}}}. 



{{{x = (-1 + 7)/(12)}}} or {{{x = (-1 - 7)/(12)}}} Break up the expression. 



{{{x = (6)/(12)}}} or {{{x =  (-8)/(12)}}} Combine like terms. 



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



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