Find m with which quadratic equation has 2 positive answers I need to find which values of $m$ will cause
$$x^2-mx-x+m+4=0$$
to have two positive solutions.
So what I know that discriminant should be positive too
$$D>0$$
$$(m+1)^2-4(m+4)>0$$
$$(m+3)(m-5)>0$$
$$m\in (-\infty;-3)\cup(5;\infty)$$
I also know that x minima should be positive, so
$$-\frac{-(m+1)}{2}>0$$
$$m>-1$$
But unfortunately it's not enough, could you help me to find solution?
 A: Assuming that when you say "two positive solutions" you mean "two positive distinct solutions" you are correct that you need the discriminant to be positive, and you've correctly computed this.
But I don't think you figured out the correct condition for the roots to be positive. You can have a quadratic with vertex in the positive numbers, but one root positive and one root negative. For example, take $(x+1)(x-4)=x^2-3x-4$. The minimum occurs at $x=\frac{3}{2}$, but one root is negative.
So while having the vertex in the positive numbers is necessary, it is not sufficient.
Instead, think about the quadratic formulas for the roots. Since the roots are
$$\frac{(m+1) \pm \sqrt{D}}{2}.$$
The smaller root will occur when we take $\frac{m+1-\sqrt{D}}{2}$, so you want
$m+1-\sqrt{D}\gt 0$, which means you want $m+1\gt \sqrt{D}$.
Edit and fix. If you square both sides, we get $(m+1)^2 \gt D = (m+1)^2 - 4(m+4)$, which means you want $0\gt -4(m+4)$, which means you want $-4(m+4)\lt 0$, or $m+4 \gt 0$. But this may add spurious solutions (because of the squaring), which occur when $m+1 \lt - \sqrt{D}\lt 0$.  So we need to take out these possibilities. These spurious solutions are introduced example when $|m+1|\geq \sqrt{D}$ and $m+1\lt 0$; that is, $(m+1)^2\geq D$ and $m+1\lt 0$.
As above, these occur when $m+4\gt 0$ and $m+1\lt 0$, that is, when $-4\lt m\lt -1$. 
So we need $m\gt -4$, and not to have $m\lt -1$. This gives $m\geq -1$ in summary.
So the conditions are:  $m\in (-\infty,-3)$ and $m\geq -1$ (impossible); or $m\in (5,\infty)$ and $m\geq -1$, which yield $m\in (5,\infty)$. 
If you are okay with a single double root which is positive, then you also allow $D=0$, which means you need to include $x=5$ into your solution set.
A: If both roots are $> 0$ then so is their sum $\rm\:m+1\:,\:$ hence $\rm\: m> -1\:,\:$ which excludes $\rm\:m\in (-4,-3)\:.$  In the remaining "real" interval $\rm\:I = (5,\infty)$ the root sum $\rm\:m+1 > 0\:,\:$ thus at least one root is $ > 0\:.\:$ But since also in $\rm\:I\:$ the root product $\rm\:m+4\: > 0\:,\:$ it follows that both roots are $> 0$ for $\rm\:m\in (5,\infty)$.
A: Quadratic equation $ax^2+bx+c=0$ 
has positive roots if
$b^2-4ac>0$
$x_1+x_2={-b\over a}>0$
$x_1*x_2={c\over a}>0$

negative roots if
$b^2-4ac>0$
$x_1+x_2={-b\over a}<0$
$x_1*x_2={c\over a}>0$

two zero roots if
$b=0$
$c=0$

has one positive root and one negative root if
$b^2-4ac>0$
$x_1*x_2={c\over a}<0$

and if you use the first statement you should get
$m\in(5,\infty)$

Regards 
MongolGenius
