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#