Question 545446


{{{3x^2+2x-4=0}}} Start with the given equation.



Notice that the quadratic {{{3x^2+2x-4}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=3}}}, {{{B=2}}}, and {{{C=-4}}}



Let's use the quadratic formula to solve for "x":



{{{x = (-B +- sqrt( B^2-4AC ))/(2A)}}} Start with the quadratic formula



{{{x = (-(2) +- sqrt( (2)^2-4(3)(-4) ))/(2(3))}}} Plug in  {{{A=3}}}, {{{B=2}}}, and {{{C=-4}}}



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



{{{x = (-2 +- sqrt( 4--48 ))/(2(3))}}} Multiply {{{4(3)(-4)}}} to get {{{-48}}}



{{{x = (-2 +- sqrt( 4+48 ))/(2(3))}}} Rewrite {{{sqrt(4--48)}}} as {{{sqrt(4+48)}}}



{{{x = (-2 +- sqrt( 52 ))/(2(3))}}} Add {{{4}}} to {{{48}}} to get {{{52}}}



{{{x = (-2 +- sqrt( 52 ))/(6)}}} Multiply {{{2}}} and {{{3}}} to get {{{6}}}. 



{{{x = (-2 +- 2*sqrt(13))/(6)}}} Simplify the square root  (note: If you need help with simplifying square roots, check out this <a href=http://www.algebra.com/algebra/homework/Radicals/simplifying-square-roots.solver> solver</a>)  



{{{x = (-2+2*sqrt(13))/(6)}}} or {{{x = (-2-2*sqrt(13))/(6)}}} Break up the expression.  



{{{x = (-1+sqrt(13))/(3)}}} or {{{x = (-1-sqrt(13))/(3)}}} Reduce



So the solutions are {{{x = (-1+sqrt(13))/(3)}}} or {{{x = (-1-sqrt(13))/(3)}}}