How to solve the equation 2x(4 - x)^-1/2 - 3(√4 - x) = 0

Please explain this Im so confused. I already know the answer is 12/5 but I have no idea how to get that answer.

Assuming you mean 3√(4 - x) instead of 3(√4 - x),
2x/√(4 - x) - 3√(4 - x) = 0
√(4 - x)[ 2x/√(4 - x) - 3√(4 - x) = 0 ]
2x - 3(4 - x) = 0
2x - 12 + 3x = 0
5x - 12 = 0
5x = 12
x = 12/5

Some questions about your notation.
Is that first term 2 times (4 - x)^(-1/2), or is that 2 times the variable x times (4 - x)^(-1/2)?
In the second term, is it just sqrt(4) or is that sqrt(4 - x)?

I'll assume the answers are "variable x" and "sqrt(4 - x)". Then you can factor out (4 - x)^(-1/2) to get

(4 - x)^(-1/2) * [2x - 3(4 - x)] = 0 because in the second term (4 - x)^(-1/2) * (4 - x) = (4 - x)^(1/2) = sqrt(4 - x).

By the factoring theorem, one of these factors must be 0. So either
(4 - x)^(-1/2) = 1/sqrt(4 - x) = 0
or
[2x - 3(4 - x)] = 0

The first one can never equal 0. No matter what x is, 1/sqrt(4 - x) is going to be a nonzero number. (A reciprocal of a real number can't ever be 0).

So that leaves us with the second one:
2x - 12 + 3x = 0
and if you solve that, you'll get x = 12/5.

...2x
------------ - 3√(4 - x) = 0
√(4 - x)

....2x
[------------- = 3√(4 - x)]√(4 - x)
√(4 - x)

2x = 3(4 - x)
2x = 12 - 3x
2x +3x = 12
5x = 12
x= 12/5 answer//