Question 318391


{{{a^2-3a-28=0}}} Start with the given equation.



Notice that the quadratic {{{a^2-3a-28}}} is in the form of {{{Aa^2+Ba+C}}} where {{{A=1}}}, {{{B=-3}}}, and {{{C=-28}}}



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



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



{{{a = (-(-3) +- sqrt( (-3)^2-4(1)(-28) ))/(2(1))}}} Plug in  {{{A=1}}}, {{{B=-3}}}, and {{{C=-28}}}



{{{a = (3 +- sqrt( (-3)^2-4(1)(-28) ))/(2(1))}}} Negate {{{-3}}} to get {{{3}}}. 



{{{a = (3 +- sqrt( 9-4(1)(-28) ))/(2(1))}}} Square {{{-3}}} to get {{{9}}}. 



{{{a = (3 +- sqrt( 9--112 ))/(2(1))}}} Multiply {{{4(1)(-28)}}} to get {{{-112}}}



{{{a = (3 +- sqrt( 9+112 ))/(2(1))}}} Rewrite {{{sqrt(9--112)}}} as {{{sqrt(9+112)}}}



{{{a = (3 +- sqrt( 121 ))/(2(1))}}} Add {{{9}}} to {{{112}}} to get {{{121}}}



{{{a = (3 +- sqrt( 121 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{a = (3 +- 11)/(2)}}} Take the square root of {{{121}}} to get {{{11}}}. 



{{{a = (3 + 11)/(2)}}} or {{{a = (3 - 11)/(2)}}} Break up the expression. 



{{{a = (14)/(2)}}} or {{{a =  (-8)/(2)}}} Combine like terms. 



{{{a = 7}}} or {{{a = -4}}} Simplify. 



So the solutions are {{{a = 7}}} or {{{a = -4}}}