How do I find the intersections for 2 non-linear (quadratic and exponential) equations without drawing or using a graphing calculator? How do I find the intersections for the equations, $y = 4-x^2$ and $y = 2^x + 1$ without drawing or using a graphing calculator?
I got this far: 
$x^2 + 2^x - 3 = 0$
Tried using some calculating programs but they could not compute the answer, and I could only find the x and y values by using a graph calculator. How can I find the x and y values without one?
 A: Continuing from  @calculus' comments, consider the function $$f(x)=x^2 + 2^x - 3$$ As noticed, $x=1$ is one solution.
On the other hand, $f(0)=-2$ and $f''(x)=2^x \log ^2(2)+2 \gt 0$; so there is another root. Using inspection $f(-1)=-\frac{3}{2}$, $f(-2)=\frac{5}{4}$ which shows a solution between $-2$ and $-1$.
Now, start Newton method, say with $x_0=-\frac{3}{2}$. This will produce as successive iterates $-1.64390$, $-1.63659$, $-1.63658$ which is the solution for six significant figures.
A: You should be able to look at the function and tell that $x=1$ is a solution. As for the other point of intersection it would be very hard by hand, but if you work with a scientific calculator, as already stated you may use Newtons method.
$$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$
$x_1$ is a guess to what might be the solution. I would guess  a negative number. As we get $x_2, x_3,x_4.....$ we get closer and closer to the answer (in terms of only $x$) you're looking for.
Note that $f'(x_n)$ is the derivative of a function, for this function:
$$f'(x_n)=2x_n+ln2(2^{x_n})$$
And so this is the formula we use to get better and better guesses: 
$$x_{n+1}=x_n-\frac{(x_n)^2+2^{x_n}-3}{2x_n+ln2(2^{x_n}) }$$
A: GeoGebra will solve it (in the CAS View) quite happily

NSolve[x^2 + 2^x - 3 = 0]
{x = -1.636576038445, x = 1}

