Give the equations that are a tangent to the parabola $y = x^2 + 5x + 6$ and pass through $(1,1)$ I have been given the question:

Give the equations that are a tangent to the parabola: $y = x^2 + 5x + 6$ and pass through the point $(1,1)$

I have tried two different methods for solving this. The first I don't know why am I incorrect (I would appreciate if someone would explain to me where I went wrong).
First Attempt:
$y = x^2 + 5x + 6$
Use $y - y_1 = m(x - x_1)$ to create lines that go through $(1,1)$
$y - 1 = m(x - 1)$
$y = mx + 1 - m$
Find when these lines intersect with the parabola:
$x^2 + 5x + 6 = mx + 1 - m$
$x^2 + x(5 - m) + (5 + m) = 0$
Find for what values of M these lines only intersect once using $b^2 - 4ac = 0$
$(5-m)^2 - 4(5+m) = 0 = m^2 - 14m + 5$
$(m - 7)^2 - 49= -5$
$m = \pm \sqrt{44} + 7$
Giving the equations:
$$
y = (\sqrt{44} + 7)x - \sqrt{44} + 1
$$
$$
y = (-\sqrt{44} + 7)x + \sqrt{44} + 1
$$
With the second attempt I am lost as to what that last equation is representing so I just got confused.
second:
$ y = x^2 + 5x + 6 $
$ y' = 2x + 5 $
All the lines that have the same gradient as a the parabola and pass through $(1,1)$
$ y - 1 = (2x + 5)(x - 1) $
$y = 2x^2 + 3x - 4$
Find when these intersect the parabola:
$x^2 + 5x + 6 = 2x^2 + 3x - 4$
which gave:
$x = \pm3 + 1$
From here I got lost and didn't know what was going on.
I would appreciate it if someone could explain why my first method is wrong, tell me if either of the methods are even on track to solving the issue and give an answer to the problem.
Thanks
-Kingpulse

Solution for method one
The first method was actually correct apart from at the very end.
The mistake was the incorrect substitution into $-m$, I was substituting $-\sqrt{44}$ into $-m$ not $-(\sqrt{44}+7)$.
Proof that it is a tangent:
Parabola : $y = x^2 + 5x + 6$
Find intersections of lines
$x^2 + 5x + 6 = (\sqrt{44} + 7)x - (\sqrt{44} + 7) + 1$
$x^2 + x(-2-\sqrt{44}) + (12 + \sqrt{44}) = 0$
$b^2 - 4ac = 0$ if it is a tangent (only one root)
$(-2-\sqrt{44})^2 - 4(12+\sqrt{44})$
$4+4\sqrt{44} + 44 - 48 -4\sqrt{44} = 0$
It is a tangent.
The equations are:
$$ y = (\sqrt{44} + 7)x - (\sqrt{44} + 7) + 1$$
and
$$ y = (-\sqrt{44} + 7)x - (-\sqrt{44} + 7) + 1$$
 A: HINT: 
For the second use parametric form $(t,t^2+5t+6)$
$$\dfrac{y-(t^2+5t+6)}{x-t}=(2x+5)_{(\text{ at }x=t)}$$
Now set $x=1,y=1$ to find the two values of $t$ to b $$1\pm\sqrt{11}$$
$$m=(2x+5)_{(\text{ at }x=t)}=2t+5=?$$
A: The first method is OK. For the second method, you have to consider a point $x_0,y_0)$  on the parabola, write the equation of the tangent to  the parabola at that point:
$$y=(2x_0+5)(x-x_0)+x_0^2+5x_0+6=(2x_0+5)x-x_0^2+6$$
and write this tangent passes though the point $(1,1)$:
$$1=2x_0+5-x_0^2+6\iff x_0^2-2x_0-10=0.$$
The reduced discriminant is $\Delta'=11$, hence 
$$x_0=1\pm\sqrt 11.$$
A: Your computation for $b^2-4ac$ should have $5+m$ in the second term. 
Also in the second method your notation is confusing you and you have $y=2x^2+3x-4$ as the equation of a straight line, which is wrong. You need to deal with the point you know, which is $(x_0,y_0)$ on the parabola, where the gradient is $2x_0+5$, work out the equation of the tangent line, and then, when you have it, find the condition that it passes through $(1,1)$.
