Use the method x = (-b +/- sqrt (b^b - 4ac) ) / 2a to solve for those imaginery roots of x^x - 4x -3.
This gives you x = 2 + sqrt7 and 2 - sqrt7
x^x - 4x + 3 can be solved by normal factorisation to (X-3)(x-1) = 0, giving x=3 and x=1
A correction to your formula ---> it is b^2, not b^b.
Thanks for the explanation!
But do problems like this really show up on the GMAT, where we have to use the quadratic formula??