Question 299581


{{{a^2-12a+35=0}}} Start with the given equation.



Notice that the quadratic {{{a^2-12a+35}}} is in the form of {{{Aa^2+Ba+C}}} where {{{A=1}}}, {{{B=-12}}}, and {{{C=35}}}



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



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



{{{a = (-(-12) +- sqrt( (-12)^2-4(1)(35) ))/(2(1))}}} Plug in  {{{A=1}}}, {{{B=-12}}}, and {{{C=35}}}



{{{a = (12 +- sqrt( (-12)^2-4(1)(35) ))/(2(1))}}} Negate {{{-12}}} to get {{{12}}}. 



{{{a = (12 +- sqrt( 144-4(1)(35) ))/(2(1))}}} Square {{{-12}}} to get {{{144}}}. 



{{{a = (12 +- sqrt( 144-140 ))/(2(1))}}} Multiply {{{4(1)(35)}}} to get {{{140}}}



{{{a = (12 +- sqrt( 4 ))/(2(1))}}} Subtract {{{140}}} from {{{144}}} to get {{{4}}}



{{{a = (12 +- sqrt( 4 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{a = (12 +- 2)/(2)}}} Take the square root of {{{4}}} to get {{{2}}}. 



{{{a = (12 + 2)/(2)}}} or {{{a = (12 - 2)/(2)}}} Break up the expression. 



{{{a = (14)/(2)}}} or {{{a =  (10)/(2)}}} Combine like terms. 



{{{a = 7}}} or {{{a = 5}}} Simplify. 



So the solutions are {{{a = 7}}} or {{{a = 5}}}