Showing that roots of a quadratic polynomial are of opposite signs.

A general quadratic with roots $\alpha$ and $\beta$ can be written $$ x^2-(\alpha+\beta)x+\alpha\beta=0 $$ The last, constant term is the product of the roots. In your case that equals $-3$. The roots must therefore have opposite signs.

In the quadratic $P(x) = x^2 - (2 + k)x - 3 = 0$, the coefficient of $x^2$ is positive. Thus, $P(X)$ and $P(-X)$ is positive for some large enough $X>0$.

But $P(0) = -3$ is negative. So between $-X$ and $0$ there is a $x_1$ where $P(x_1) = 0$ and between $0$ and $X$ there is a $x_2$ where $P(x_2) = 0$: the polynomial has real roots and they are of opposite signs.

Well, the solutions to the equation are

$$x=\frac{2+k}{2}\pm\frac12\sqrt{(-2-k)^2-4\cdot1\cdot-3}$$

or, simplified,

$$x=\frac{2+k}2\pm\frac12\sqrt{(2+k)^2+12}$$

and this is equal to

$$x=\frac{2+k}2\pm\sqrt{\left(\frac{2+k}2\right)^2+3}$$

The important step here is

$$\sqrt{\left(\frac{2+k}2\right)^2+3}>\sqrt{\left(\frac{2+k}2\right)^2}=\left|\frac{2+k}2\right|$$

and so if we choose $+$, then we see the root is positive:

$$\frac{2+k}2+\sqrt{\left(\frac{2+k}2\right)^2+3}>\frac{2+k}2+\left|\frac{2+k}2\right|\geq0$$

When we choose $-$, we get something negative:

$$\frac{2+k}2-\sqrt{\left(\frac{2+k}2\right)^2+3}<\frac{2+k}2-\left|\frac{2+k}2\right|\leq0$$

thus, the roots must be of different signs.