Solving $x^2+ \ln^2(x)=1$

First of all, this isn't a quadratic. Secondly, this is a type of an equation that cannot be solved explicitly, that is, to find a closed form expression for $x$. The only way to work such equations around is using a Numerical Analysis approach with a preffered method.

Apart from that, indeed one can see that $x=1$ trully satisfies the equation $x^2 + \ln^2 x - 1 = 0$. One could study the behavior of the function $f(x)$ yielded by that expression, as to know where the other solution is around.

Let $f(x) = x^2 + \ln^2 x - 1$. Then, $f'(x) = 2x - \frac{2\ln x}{x}$. Now, investigate :

$$f'(x) = 0 \Rightarrow 2x - \frac{2 \ln x}{x} = 0 \Rightarrow 2x^2 - 2\ln x =0$$

We've again come across a similar, non-explicitly solvable equation. Working now, as :

$$f''(x) = 0 \Rightarrow 4x - \frac{2}{x} = 0 \Rightarrow 4x^2 - 2 = 0 \Leftrightarrow x = \frac{\sqrt{2}}{2}$$

Note that we excluded the negative part of the solution, as $x >0$ since we are working with a logarithm expression. That, now, can give you a hint on where to look for the other solution and thus how to manipulate/start your preffered numerical method.