Question 549931


{{{x^2-13x-12=0}}} Start with the given equation.



Notice that the quadratic {{{x^2-13x-12}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=1}}}, {{{B=-13}}}, and {{{C=-12}}}



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



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



{{{x = (-(-13) +- sqrt( (-13)^2-4(1)(-12) ))/(2(1))}}} Plug in  {{{A=1}}}, {{{B=-13}}}, and {{{C=-12}}}



{{{x = (13 +- sqrt( (-13)^2-4(1)(-12) ))/(2(1))}}} Negate {{{-13}}} to get {{{13}}}. 



{{{x = (13 +- sqrt( 169-4(1)(-12) ))/(2(1))}}} Square {{{-13}}} to get {{{169}}}. 



{{{x = (13 +- sqrt( 169--48 ))/(2(1))}}} Multiply {{{4(1)(-12)}}} to get {{{-48}}}



{{{x = (13 +- sqrt( 169+48 ))/(2(1))}}} Rewrite {{{sqrt(169--48)}}} as {{{sqrt(169+48)}}}



{{{x = (13 +- sqrt( 217 ))/(2(1))}}} Add {{{169}}} to {{{48}}} to get {{{217}}}



{{{x = (13 +- sqrt( 217 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{x = (13+sqrt(217))/(2)}}} or {{{x = (13-sqrt(217))/(2)}}} Break up the expression.  



So the solutions are {{{x = (13+sqrt(217))/(2)}}} or {{{x = (13-sqrt(217))/(2)}}}