Find $m$ such that the roots of this polynomial are greater than -1. The polynomial $x^2 + 2mx + 3m+4$. 
I know that the discrimate must be greater or equal to zero, otherwise it would have complex roots, and complex numbers are "measureable", don't know how else to explain, against -1.  Because complex numbers have two parts, while -1 is doesn't.  
So, the discrimate $b^2-4ac \ge -1$ results with $(m+1)(m-4)\ge -1$ means that either $m \le -1 \cup m \ge 4$ 
Now, I am stuck upon what comes next. Please, help. 
 A: May be you shuld look at the following way :
roots of $x^2+2mx+3m+4=0$ are $\frac{-2m\pm \sqrt{4m^2-12m-16}}{2}$
You want $\frac{-2m\pm \sqrt{4m^2-12m-16}}{2}>-1$ i.e., $-2m\pm \sqrt{4m^2-12m-16}>-2$
i.e., $\sqrt{4m^2-12m-16}> 2(m-1)$
I guess you know how to continue from this....
A: Observation: $a,b>0\iff a+b,ab>0$. Why?
Consider further: $a,b>-1\iff a+1,b+1>0$.
Suppose $(x-a)(x-b)=x^2+Ux+V$. Then what is $\big(x-(a+1)\big)\big(x-(b+1)\big)$? Compute
$$\begin{array}{ll} (x-(a+1))(x-(b+1)) & =x^2-(a+b+2)x+(ab+a+b+1) \\ & =x^2+(U-2)x+(V-U+1). \end{array} $$
Corollary. The roots of $x^2+Ux+V$ (presumed real) are $>-1$ iff $U<2$ and $V-U+1>0$.
Apply with $U=2m$ and $V=3m+4$ while setting the discriminant $\Delta\ge0$.
A: Not sure how much resemblance it has with anon's answer
Let $x_1,x_2$ are the two roots of $x^2 + 2mx + 3m+4=0\ \ \ \ (1)$ 
Let us find the equation whose roots are $x_1+1=y_1>0,x_2+1=y_2>0$ 
As $y_1-1=x_1$ and $x_1$ satisfies $\displaystyle(1)\implies (y_1-1)^2 + 2m(y_1-1) + 3m+4=0$
$\displaystyle\iff y_1^2+2y_1(m-1)+m+5=0$ 
Similarly,  $\displaystyle y_2^2+2y_2(m-1)+m+5=0$ 
So, $y_1,y_2>0,$ are the roots of $\displaystyle y^2+2y(m-1)+m+5=0\  \ \  \ (2)$
First of all, the discriminant must be $\ge0\  \ \  \ (3)$
$\displaystyle y_1y_2=m+5$ which must be $>0\  \ \  \ (4)$
$\displaystyle y_1+y_2=-\frac{2(m-1)}1$ which must be $>0\  \ \  \ (5)$
Find the intersection of $(3),(4),(5)$
