Question 228192

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



Notice that the quadratic {{{3x^2+x-2}}} is 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( 1-4(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 = (4)/(6)}}} or {{{x =  (-6)/(6)}}} Combine like terms. 



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



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