Algebra
posted by Sam .
How do I find x for this equation? I think I should get e^(something).
Thanks in advance!
3x^2+4lnx+4=0

There's no algebraic way to do it. A numeric method or a graphical analysis will be required. See
http://www.wolframalpha.com/input/?i=solve+3x^2%2B4lnx%2B4%3D0 
For a complicated equation such as yours, the most powerful method is probably Newton's Method.
let y = 3x^2+4lnx+4
dy/dx = 6x + 4/x
newx = x  y/(dy/dx)
= x  (3x^2+4lnx+4)/(6x + 4/x)
which simplifies to
newx = (3x^2  4lnx)/(6x + 4/x)
now use an iteration process, I started with x = 1
x  newx
1  .3
.3  .336072..
.336072  .337716
.337716  .337719143
.337719143  .337719143
since the last x used returned the same value of x, we cannot improve on that answer.
Amazing that after only 4 interations, I got
x =.337719143 correct to 9 decimal places