Question 598378


{{{x^2-x-15=0}}} Start with the given equation.



Notice that the quadratic {{{x^2-x-15}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=1}}}, {{{B=-1}}}, and {{{C=-15}}}



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



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



{{{x = (-(-1) +- sqrt( (-1)^2-4(1)(-15) ))/(2(1))}}} Plug in  {{{A=1}}}, {{{B=-1}}}, and {{{C=-15}}}



{{{x = (1 +- sqrt( (-1)^2-4(1)(-15) ))/(2(1))}}} Negate {{{-1}}} to get {{{1}}}. 



{{{x = (1 +- sqrt( 1-4(1)(-15) ))/(2(1))}}} Square {{{-1}}} to get {{{1}}}. 



{{{x = (1 +- sqrt( 1--60 ))/(2(1))}}} Multiply {{{4(1)(-15)}}} to get {{{-60}}}



{{{x = (1 +- sqrt( 1+60 ))/(2(1))}}} Rewrite {{{sqrt(1--60)}}} as {{{sqrt(1+60)}}}



{{{x = (1 +- sqrt( 61 ))/(2(1))}}} Add {{{1}}} to {{{60}}} to get {{{61}}}



{{{x = (1 +- sqrt( 61 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{x = (1+sqrt(61))/(2)}}} or {{{x = (1-sqrt(61))/(2)}}} Break up the expression.  



So the solutions are {{{x = (1+sqrt(61))/(2)}}} or {{{x = (1-sqrt(61))/(2)}}} 



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