Great Quadratic Equation

$x_1^2 + \sqrt{x_2^2 - 2x_2} = 2x_1 - 1 \Rightarrow (x_1 - 1)^2 = -\sqrt{x_2(x_2 - 2)}$

Notice that $(x_1 -1)^2 \geq 0$ and $-\sqrt{x_2(x_2 - 2)} \leq 0$

So, this will force to deduce that $(x_1 -1)^2 = 0$ and $-\sqrt{x_2(x_2 - 2)} = 0$

For Equation 1

$(x_1 -1)^2 = 0 \Rightarrow x_1 = 1$ .

For Equation 2

$-\sqrt{x_2(x_2 - 2)} = 0 \Rightarrow x_2 = 0$ or $x_2 = 2$

But $x_2$ is positive real solution $\Rightarrow x_2 =2$ .

Thus, $x^2 -bx+c = (x-1)(x-2) = 0$

$\Rightarrow x^2 - 3x + 2 = 0$

$\Rightarrow b = 3$ and $c = 2$

$\Rightarrow b+c = 3+2 = 5 \Rightarrow \min(b+c) = 5$