Log6(x) + log6(x+3) = 2 | help, please

There are two potential roots, A and B, where A less than/equal to B .

A=____ B=_____

Is A actually a root? (yes or no)
Is B actually a root? (yes or no)

i dont understand what is meant by A and B, i got 1 answer :/

log6(x) + log6(x+3) = 2
log6[x(x+3)] = 2
x(x+3) = 6² = 36
x² + 3x = 36
x² + 3x + 9/4 = 36+9/4 = 153/4
(x+3/2)² = ±√(153/4)
x = -3/2±√153/2
A = -(3+√153)/2 ≈ -7.66441
B = (-3+√153)/2 ≈ 4.68466

A is not a root since A < 0 and logarithms of non-positive numbers are not defined.
B is a root which can be proven by substitution of B into the original equation.

First simplify the equation

Next solve the equation

Last check if the roots work.

Simplify
log6(x) + log6(x+3) = 2 ------> combine the logs
log6 x(x+3) = 2 -------> eliminate the log
6² = x(x+3)
36 = x² + 3x

Solve the equation using the quadratic formula or complete the square method

x²+3x-36 = 0

You will get two answers
Call one A and the other B

log6(x) + log6(x + 3) = 2

log6 (x(x + 3)) = 2

x(x + 3) = 36

x^2 + 3x - 36 = 0

x = (-3 +/- √(9 - 4(-36))/2 = (-3 +/- √153)/2

Go on from here.