I think you may have typed this in wrongly because the term 2^x is not a permitted term in a quadratic equation.
As it stands the solution to the equation is about -3.6168.
Let's say you made a mistake in typing and the middle term is 2*x or 2x, then the solution exists but it is irrational (involves square roots). The easiest way to solve x^2-2x-13=0 is to complete the square:
(x^2-2x+1)-1-13=0, which is (x-1)^2-14=0, so (x-1)^2=14. Take square roots of each side: x-1=+sqrt(14)=+3.7417 approx. A square root always has two solutions, one positive and one negative, but the same magnitude number. So x has two possible values: x=1+3.7417=4.7417 or x=1-3.7417=-2.7417.
If you meant the question to be x^2-12x-13=0, then the solution is (x-13)(x+1)=0 and x=13 or -1. To work this out we ask: what are the factors of 1 (x^2 coefficient) and 13 (the constant term). The factors of 1 are 1 and 1 because only 1 times 1 make 1; and the factors of 13 are only 1 and 13. The coefficient of the x^2 term tells is how many x's go in each bracket. That's 1x or just x in each bracket. And the factors of 13 tell us what. Numbers to write in each bracket, so that's 1 and 13. So we have (x 1)(x 13). What about the signs between them? We look at the sign in front of 13 in the quadratic. It's minus, and that means there will be a plus in one bracket and a minus in the other. But which way round? Well, there's one more test: we take the factors of 13 and subtract them because the minus sign in front of 13 tells us we need to subtract. If it had been plus, we would have added the factors. 13-1=12. If 12 is the coefficient of the x term then the quadratic can be solved. (If the number had not been 12 we could not have solved the quadratic this way.) The sign in front of 12x is the sign that goes in front of the larger number in the brackets, so minus goes in front of 13.
So we have (x+1)(x-13)=0. One or other of these factors is zero, so x+1 or x-13 is zero. x+1=0 means x=-1 and x-13=0 means x=13. These are the solutions or roots.
