How do you solve #-abs(x+1)=-2#?

The answer is:
#x=1# and
#x=-3#

The way to solve it is as follows.

Equal parts of an equation, left and right, can be multiplied by the same non-equal to zero multiplier getting an equivalent equation.
Let's multiply them by #-1#:
#|x+1|=2#

Now we have to remember the definition of the absolute value of a number.
If the number is positive or zero, its absolute value equals to itself:
if #Z>=0# then #|Z|=Z#.
If the number is negative, its absolute value equals to its opposite (or, not very scientifically, minus this number)
if #Z<0# then #|Z|=-Z#.

Applying this to a problem at hand:

CASE 1
Looking for solutions in the area defined by an inequality
#x+1>=0# (that is, #x>=-1#) then #|x+1|=x+1# and,
using our equation,
#x+1=2#, that is, #x=1#.
This value is within the area #x>=-1# and, therefore, is the first legitimate solution.

CASE 2
Looking for solutions in the area defined by an inequality
#x+1<0# (that is, #x<-1#) then #|x+1|=-(x+1)# and,
using our equation,
#-(x+1)=2#, that is, #x=-3#.
This value is within the area #x<-1# and, therefore, is the second legitimate solution.

We can demonstrate it graphically.
Our equation is equivalent to
#|x+1|-2=0#
Let's draw a graph of a function
#y=|x+1|-2#
graph{|x+1|-2 [-10, 10, -5, 5]}
As you see, it intersects the X-axis (that is, equals to zero) at points #x=-3# and #x=1#. This confirms our solutions.