We look forward to more "radical questions" ...

Don't forget that a square root is just a half power:

√x = x^(1/2)

mkay!

To start, let me show you a bit of "logic":
If A=B then A²=B² (A=B => A²=B²)
eg:
-2=-2
(-2)²= (-2)²
4=4

The opposite not always is true (that's why the 'implies =>' symbol instead of the 'equivalent to <=>'):
A²=B² ≠> A=B
e.g.

A²=B²
4=4
2²=2²
2=2
A=B

A²=B²
16=16
(-4)²=4²
-4≠4
A≠B

This type of equations do not always return valid roots!
So how to proceed in order to ensure that all solutions are valid?

It's easy but very important!
-> When you'r done solving the equation, you must plug in ALL solutions into the original equation and verify if you get an equality (1=1,  x=x,... something=something)
If that does not happen you must discard that solution.

*Lets check your equations and then tryout a more complex ones*
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Eq.1
-----------------------------------------------------------------------
2√(x+4)=3
√(x+4)=3/2     'A=B => A²=B²
x+4= (3/2)²
x=9/4-4=-7/4

verification:
2√(-7/4+4)=3
2*1.5=3
3=3         'TRUE
the first eq. has solution x=-7/4

-----------------------------------------------------------------------
Eq.2
-----------------------------------------------------------------------
√(x+3)=2√(x)
x+3=2²x
x+3=4x
3=4x-x
3=3x
x=1

verification:
√(1+3)=2√(1)
√(4)=2
2=2             'TRUE
the second eq. has solution x=1

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Let try an equation were some solutions really aren't..
-----------------------------------------------------------------------
√(x+1)=x-1
x+1=(x-1)²
x+1=x²-2x+1
x=x²-2x
3x-x²=0
(3-x)x=0
x=3 or x=0

Lets verify both (supposed) solutions
x=3
√(3+1)=3-1
√(4)=2
2=2     'TRUE
great! x=3 is a solution!

Now for x=0
√(0+1)=0-1
√(1)=-1
1=-1     FALSE!!!
X=0 ISN'T A ROOT(SOLUTION) OF √(x+1)=x-1

See how important it is to verify your solutions?
I hope it was all very clear.

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All the above stuff has to do also with the domains os the √() operation in IR.
√[u(x)]=v(x) => u(x)≥0 ...we cant have sqrt of neg. numbers
You can use domains in order to validate solutions but I think that substituition is the faster & simplier method. You choose.
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&btw...please feel free to register in the forum!

Is this solution correct?

1.
2[SQRT(x+4)]=3
2[SQRT(x+4)]^2=3^2
4x+16=9
4x=9-16
4x=-7
x=-7/4

2.
[SQRT(x+3)]=2[SQRT(x)]
[SQRT(x+3)]^2=2[SQRT(x)]^2
x+9=4x I'm not sure what to do here?