Where am I going wrong while trying to solve for $m$ in this quadratic equation? Find $m$ so that the entire graph of the following function would be above the $x$ axis: 
$$y=mx^{ 2 }+(m+7)x+m+15$$
Steps I took:
The graph will be above the $x$ axis when there are no real roots, such that $b^2-4ac<0$
So I got, 
$$(m+7)^2 -4(m)(m+15)<0\Rightarrow m^2+14m+49-4m^2-60m<0\Rightarrow -3m^2-46m+49<0$$
Then I used the quadratic formula:
$$\frac { 46\pm \sqrt { (-46)^{ 2 }-4(-3)(49) }  }{ 2(-3) } \Rightarrow \frac { 46\pm \sqrt { 2704 }  }{ -6 } \Rightarrow \frac { 46\pm 52 }{ -6 } $$
I used the two solutions to turn this into: 
$$(m-1)(m+\frac { 49 }{ 3 } )<0$$
So, in order to be negative the solutions must come from: 
$$(m-1)<0,\quad (m+\frac { 49 }{ 3 } )<0,\quad (m-1)>0,\quad (m+\frac { 49 }{ 3 } )>0$$
However, the given solution $(1,\infty )$ makes no sense given the solutions I came up with. Where am I going wrong? 
 A: Here is where you are making a mistake:  The condition
$$-3m^2-46m+49<0$$
is not equivalent to the condition
$$(m-1)(m+\frac{49}{3})<0$$
even though you have correctly found the zeroes of the first polynomial.  You can check that they are not the same by multiplying out $(m-1)(m+\frac{49}{3})$ and seeing that it does not give you $-3m^2-46m+49$.
What went wrong?  Simple:  You're missing a factor of $-3$.  The condition you want is
$$-3(m-1)(m+\frac{49}{3})<0$$
But there is a simpler way to finish this that does not require a correct factorization.  Remember that the graph of $y=-3m^2-46m+49$ is a downward-pointing parabola.  Since it crosses the $x$-axis at two points, that parabola must take on positive values between the zeroes, and negative values everywhere else.
A: Hint #1:  $-3m^2 - 46m + 49 = -(3m^2 + 46m - 49)$.
Hint #2:  Multiplying both sides of an inequality by a negative number reverses the direction of the inequality.
Can you take it from here?
A: The discriminant of the right-hand side must be negative, that's right.
Now the discriminant is
$$
(m+7)^2-4m(m+15)=m^2+14m+49-3m^2-60m=-3m^2-46m+49
$$
and the inequality is
$$
-3m^2-46m+49<0
$$
that is,
$$
3m^2+46m-49>0
$$
The factorization is good:
$$
3(m-1)\left(m+\frac{49}{3}\right)>0
$$
which is satisfied when both factors are positive or both are negative, so
$$
m<-\frac{49}{3}\qquad\text{or}\qquad m>1
$$
However, what shape does the parabola have when $m<0$?
