[Incomplete question. I found the equation on other source:
√(-3X-2)=X+2]
Let’s find x:
√(-3X-2)=X+2
-3X-2=(X+2)²
-3X-2=x²+2*2*x+2²
-3X-2=x²+4x+4
-3X-2-x²-4x-4=0
-x²-7x-6=0
So now you should use Bhaskara formula. Unless your calculator makes for itself (some calculators do).
ax²+bx+c=0
x=(-b+-√(b²-4ac))/(2a)
a=-1
b=-7
c=-6
x=-1
or x=-6
Let’s verify -1√(-3*(-1)-2)=-1+2
√(3-2)=1
√1=1
1=1 TRUE -> So -1 is a solution, not an extraneous solution
Let’s verify x=-6:
√(-3*(-6)-2)=-6+2
√(18-2)=-4
√16=-4
4=-4 FALSE –> So -6 comes from the equation but does not verify. So -6 is an extraneous solution. This is related to the fact that square roots have only 1 solution, whereas squares have 2 solutions (remember (-1)² is 1, and 1² is 1 as well, because of the rules of signs for multiplication).
