Question 242738


{{{3x^2+14x-24=0}}} Start with the given equation.



Notice that the quadratic {{{3x^2+14x-24}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=3}}}, {{{B=14}}}, and {{{C=-24}}}



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



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



{{{x = (-(14) +- sqrt( (14)^2-4(3)(-24) ))/(2(3))}}} Plug in  {{{A=3}}}, {{{B=14}}}, and {{{C=-24}}}



{{{x = (-14 +- sqrt( 196-4(3)(-24) ))/(2(3))}}} Square {{{14}}} to get {{{196}}}. 



{{{x = (-14 +- sqrt( 196--288 ))/(2(3))}}} Multiply {{{4(3)(-24)}}} to get {{{-288}}}



{{{x = (-14 +- sqrt( 196+288 ))/(2(3))}}} Rewrite {{{sqrt(196--288)}}} as {{{sqrt(196+288)}}}



{{{x = (-14 +- sqrt( 484 ))/(2(3))}}} Add {{{196}}} to {{{288}}} to get {{{484}}}



{{{x = (-14 +- sqrt( 484 ))/(6)}}} Multiply {{{2}}} and {{{3}}} to get {{{6}}}. 



{{{x = (-14 +- 22)/(6)}}} Take the square root of {{{484}}} to get {{{22}}}. 



{{{x = (-14 + 22)/(6)}}} or {{{x = (-14 - 22)/(6)}}} Break up the expression. 



{{{x = (8)/(6)}}} or {{{x =  (-36)/(6)}}} Combine like terms. 



{{{x = 4/3}}} or {{{x = -6}}} Simplify. 



So the solutions are {{{x = 4/3}}} or {{{x = -6}}}