Question: I have to find the values of the real parameter $k$ for which the equation: 
$$x^2-7x+2k+3=0$$
has roots with different signs.
We need the condition $x\_1x\_2<0$ for the roots to be with different signs. Using Vieta's formulas: $x\_1+x\_2=-\frac{b}{a}=-\frac{-7}{1}=7$ and $x\_1x\_2=\frac{c}{a}=\frac{2k+3}{1}=2k+3.$ As I said $x\_1x\_2<0$, so $2k+3<0, k<-\frac{3}{2}$. Is this the correct solution and should I write the first Vieta's formula ($x\_1+x\_2=-\frac{b}{a}$)?

Answer: 
Just don't forget the discriminant should be positive to ensure the existence of roots, so:$$49-4(2k+3)>0$$
It is east to overthink and overcomplicate this.

If you sketch the function, the parameter $k$ represents a vertical shift. So you want to identify the value of $k$ which gives $x=0$ as a solution and then if you drop the curve below that (shift it down) you will have roots with both signs.

ie the condition reduces to $f(0)\lt 0$ for $f(x)=x^2-7x+2k+3$ (note that the coefficient of $x^2$ is positive)

and that is what you get from the vieta product formula - you don't need the other one.