Question 299539


{{{3x^2-11x=20}}} Start with the given equation.



{{{3x^2-11x-20=0}}} Subtract 20 from both sides.



Notice that the quadratic {{{3x^2-11x-20}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=3}}}, {{{B=-11}}}, and {{{C=-20}}}



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



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



{{{x = (-(-11) +- sqrt( (-11)^2-4(3)(-20) ))/(2(3))}}} Plug in  {{{A=3}}}, {{{B=-11}}}, and {{{C=-20}}}



{{{x = (11 +- sqrt( (-11)^2-4(3)(-20) ))/(2(3))}}} Negate {{{-11}}} to get {{{11}}}. 



{{{x = (11 +- sqrt( 121-4(3)(-20) ))/(2(3))}}} Square {{{-11}}} to get {{{121}}}. 



{{{x = (11 +- sqrt( 121--240 ))/(2(3))}}} Multiply {{{4(3)(-20)}}} to get {{{-240}}}



{{{x = (11 +- sqrt( 121+240 ))/(2(3))}}} Rewrite {{{sqrt(121--240)}}} as {{{sqrt(121+240)}}}



{{{x = (11 +- sqrt( 361 ))/(2(3))}}} Add {{{121}}} to {{{240}}} to get {{{361}}}



{{{x = (11 +- sqrt( 361 ))/(6)}}} Multiply {{{2}}} and {{{3}}} to get {{{6}}}. 



{{{x = (11 +- 19)/(6)}}} Take the square root of {{{361}}} to get {{{19}}}. 



{{{x = (11 + 19)/(6)}}} or {{{x = (11 - 19)/(6)}}} Break up the expression. 



{{{x = (30)/(6)}}} or {{{x =  (-8)/(6)}}} Combine like terms. 



{{{x = 5}}} or {{{x = -4/3}}} Simplify. 



So the solutions are {{{x = 5}}} or {{{x = -4/3}}}