Question 200869
{{{a(a+1)-3(a+1)=a+1}}} Start with the given equation.



{{{a^2+a-3a-3=a+1}}} Distribute



{{{a^2+a-3a-3-a=1}}} Subtract "a" from both sides.



{{{a^2+a-3a-3-a-1=0}}} Subtract 1 from both sides.



{{{a^2-3a-4=0}}} Combine like terms.



Notice that the quadratic {{{a^2-3a-4}}} is in the form of {{{Aa^2+Ba+C}}} where {{{A=1}}}, {{{B=-3}}}, and {{{C=-4}}}



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)(-4) ))/(2(1))}}} Plug in  {{{A=1}}}, {{{B=-3}}}, and {{{C=-4}}}



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



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



{{{a = (3 +- sqrt( 9--16 ))/(2(1))}}} Multiply {{{4(1)(-4)}}} to get {{{-16}}}



{{{a = (3 +- sqrt( 9+16 ))/(2(1))}}} Rewrite {{{sqrt(9--16)}}} as {{{sqrt(9+16)}}}



{{{a = (3 +- sqrt( 25 ))/(2(1))}}} Add {{{9}}} to {{{16}}} to get {{{25}}}



{{{a = (3 +- sqrt( 25 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{a = (3 +- 5)/(2)}}} Take the square root of {{{25}}} to get {{{5}}}. 



{{{a = (3 + 5)/(2)}}} or {{{a = (3 - 5)/(2)}}} Break up the expression. 



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



{{{a = 4}}} or {{{a = -1}}} Simplify. 



So the solutions are {{{a = 4}}} or {{{a = -1}}}