Question 472582


{{{3k^2+7k=4}}} Start with the given equation.



{{{3k^2+7k-4=0}}} Subtract 4 from both sides.



Notice that the quadratic {{{3k^2+7k-4}}} is in the form of {{{Ak^2+Bk+C}}} where {{{A=3}}}, {{{B=7}}}, and {{{C=-4}}}



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



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



{{{k = (-(7) +- sqrt( (7)^2-4(3)(-4) ))/(2(3))}}} Plug in  {{{A=3}}}, {{{B=7}}}, and {{{C=-4}}}



{{{k = (-7 +- sqrt( 49-4(3)(-4) ))/(2(3))}}} Square {{{7}}} to get {{{49}}}. 



{{{k = (-7 +- sqrt( 49--48 ))/(2(3))}}} Multiply {{{4(3)(-4)}}} to get {{{-48}}}



{{{k = (-7 +- sqrt( 49+48 ))/(2(3))}}} Rewrite {{{sqrt(49--48)}}} as {{{sqrt(49+48)}}}



{{{k = (-7 +- sqrt( 97 ))/(2(3))}}} Add {{{49}}} to {{{48}}} to get {{{97}}}



{{{k = (-7 +- sqrt( 97 ))/(6)}}} Multiply {{{2}}} and {{{3}}} to get {{{6}}}. 



{{{k = (-7+sqrt(97))/(6)}}} or {{{k = (-7-sqrt(97))/(6)}}} Break up the expression.  



So the solutions are {{{k = (-7+sqrt(97))/(6)}}} or {{{k = (-7-sqrt(97))/(6)}}}