Real roots
This is a quadratic in e^x. It will be easier to see what we're doing if we introduce a substitution. I will let t=e^x:
(e^x)^2-6e^x-16=0
t^2-6t-16=0
We solve the quadratic by grouping:
t^2-8t+2t-16=0
t(t-8)+2(t-8)=0
(t+2)(t-8)=0
By the zero factor principle, this has solutions t=-2 and t=8. We can undo the substitution to get these in terms of x:
e^x=8
ln(e^x)=ln(8)
x=ln(8)
Complex roots
For the other root, t=-2, we can try doing the same procedure:
e^x=-2
ln(e^x)=ln(-2)
x=ln(-2)
But we run into a problem. The natural log of negative numbers is not defined with real numbers, which means that this root is complex.
We can find this root by using Euler's identity, e^(ipi)=-1. This lets us rewrite the logarithm as such:
ln(-2)=ln(2*-1)=ln(2e^(ipi))=ln(2)+ln(e^(ipi))=ln(2)+ipi
Note that ipi is not the only imaginary exponent that gives -1. Euler's formula actually gives an infinite number of ways to express -1 in this way due to the periodicity of the sine and cosine functions. We can capture this by adding 2pik (remember, the periodicity of sine and cosine is 2pi) in the exponent, which after the same calculations as before gives the solutions:
x=ln(2)+(2k+1)ipi
Using complex numbers, we can also find even more solutions!
Because Euler's identity gives that e^(i2pik)=1, we can do the same thing we did with ln(-2) with ln(8):
ln(8)=ln(8e^(i2pik))=ln(8)+ln(e^(i2pik))=ln(8)+i2pik
