Solve $x^2+x^2t^2+2xt^2+t^2=1$ for $x$. Solving
$$x^2+x^2t^2+2xt^2+t^2=1$$
for $x$ yields
$$x=\frac{1-t^2}{t^2+1}.$$
I had to use software because I could not manipulate it algebraically, and sadly the program did not tell me how it reached that solution.
What is the method?
 A: This is a quadratic equation in $x$. Use the Quadratic Formula.
Recall that the solutions of $ax^2+bx+c=0$ are $\frac{-b\pm\sqrt{b^2-4ac}}{2a}$.
When you do the details, you will find that $x=-1$ is also a solution.
Another way: In this case, we can also use factorization. The polynomial 
$$x^2(1+t^2)+2xt^2+t^2-1$$
factors as
$$(x+1)\left((1+t^2)x+(t^2-1)\right),$$
and then the roots are easy to pick up.
Still another way: By inspection, $x=-1$ is a root. But the product of the roots of $ax^2+bx+c=0$ is $\frac{c}{a}$. In our case, the product of the roots is $\frac{t^2-1}{1+t^2}$. Since one of the roots is $-1$, the other one must be $\frac{1-t^2}{1+t^2}$.
Remark: My guess is that you are solving the equation because you are trying to find where the line through $(-1,0)$ with slope $t$ meets the unit circle. If that is the case, then the solution $x=-1$ is obvious even without looking at the equation. Thus the "Still another way" is the most natural way to proceed: If you have a quadratic, and you know one root, the other root requires almost no work. Just use the fact that the product of the roots is $\frac{c}{a}$ or the fact that the sum of the roots is $-\frac{b}{a}$. 
A: Collect the coefficients in $x$ to get a polynomial in $x$:
$$ (t^2+1) x^2 +(2t^2)x+(t^2-1) = 0 $$
Now solve using quadratic formula:
$$ \frac{-2t^2 \pm \sqrt{(2t^2)^2 - 4(t^2+1)(t^2-1)}}{2(t^2+1)} = \\
 \frac{-2t^2 \pm 2}{2(t^2+1)} = \frac{-t^2 \pm 1}{t^2 + 1}$$
So either $x = -1$ or $x = \frac{-t^2 + 1}{t^2 + 1}$.
A: You got a great hint earlier, but you might want to also look at the implicit plot and see another hidden fact.
See, for example:
http://www.wolframalpha.com/input/?i=Solve%5Bx%5E2%2Bx%5E2+t%5E2+%2B+2+x+t%5E2+%2B+t%5E2+%3D%3D+1%2C+x%5D
A: This equation can be written as  $x^2+t^2(x+1)^2=1$ which $\implies t^2=\frac{(1+x)(1-x)}{(1+x)^2}\implies t^2=\frac{(1-x)}{(1+x)}$ (assuming $x\neq -1$) $ \implies t^2(1+x)=1-x\implies t^2+t^2x=1-x\implies x(t^2+1)=1-t^2\implies x=\frac{1-t^2}{1+t^2}$
Here, we have to check whether $x=-1$ satisfies the equation separately. Putting in the equation $x=-1$, it satisfies the equation. Thus  $x=-1, \frac{1-t^2}{1+t^2}$  
