Question 592841


{{{3x^2+8x-24=0}}} Start with the given equation.



Notice that the quadratic {{{3x^2+8x-24}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=3}}}, {{{B=8}}}, 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 = (-(8) +- sqrt( (8)^2-4(3)(-24) ))/(2(3))}}} Plug in  {{{A=3}}}, {{{B=8}}}, and {{{C=-24}}}



{{{x = (-8 +- sqrt( 64-4(3)(-24) ))/(2(3))}}} Square {{{8}}} to get {{{64}}}. 



{{{x = (-8 +- sqrt( 64--288 ))/(2(3))}}} Multiply {{{4(3)(-24)}}} to get {{{-288}}}



{{{x = (-8 +- sqrt( 64+288 ))/(2(3))}}} Rewrite {{{sqrt(64--288)}}} as {{{sqrt(64+288)}}}



{{{x = (-8 +- sqrt( 352 ))/(2(3))}}} Add {{{64}}} to {{{288}}} to get {{{352}}}



{{{x = (-8 +- sqrt( 352 ))/(6)}}} Multiply {{{2}}} and {{{3}}} to get {{{6}}}. 



{{{x = (-8 +- 4*sqrt(22))/(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 = (-8+4*sqrt(22))/(6)}}} or {{{x = (-8-4*sqrt(22))/(6)}}} Break up the expression.



{{{x = (-4+2*sqrt(22))/(3)}}} or {{{x = (-4-2*sqrt(22))/(3)}}} Reduce



So the solutions are {{{x = (-4+2*sqrt(22))/(3)}}} or {{{x = (-4-2*sqrt(22))/(3)}}}