if you rewrite this equation to:
#a^2-144a+24=0#
then
a=1 (it's the number in fron of the number #a^2#)
b=-144
c=24
The well known formula:
#x_(1|2)=(-b+-sqrt(b^2-4ac))/(2a)#
in our case we are finding the #a# so it should be a bit different (instead of a letter x the should be a) but then it gets a bit confusing. So let's just substitute those number: a,b,c and find its value
#x_(1|2)=(-(-144)+-sqrt((-144)^2-4*1*24))/(2*1)#
with the help of calculator we get this
#x_(1|2)=(144+-sqrt(20736-96))/(2)#
#x_(1|2)=(144+-sqrt(20736-96))/(2)#
that is approximately
#x_(1|2)~~(144+-(143.66))/(2)#
When I said that we are looking for #a# as it stands in the top so don't mind that there is #x_(1|2)=# it's the same just written a bit different. so
#a_1~~0.34/2=0.17#
#a_2~~287.66/2=143.83#
