Question 470144
I'm going to use the variable x instead of 'a'.



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



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



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



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



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



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



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



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



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



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



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



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



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



{{{x = 4}}} or {{{x = -7}}} Reduce. 



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