Simple factorization problem I cannot find the solution! This is not a homework assignment. 

$11x^2+13x-7$

 A: Oh boy, do I have some good news for you. Allow me to introduce you to the math world's latest invention:
$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
The quadratic formula!
Basically, your equation is quadratic (the highest exponent on an $x$ is $2$), and it has the general form:
$$y=ax^2+bx+c$$
where $$a=11,b=13,c=-7$$
Plugging these values into the equation and evaluating gives two solutions:
$$x_{1}=\frac{-b+\sqrt{b^2-4ac}}{2a}\quad\text{AND}\quad x_{2}=\frac{-b-\sqrt{b^2-4ac}}{2a}$$
Now you can finish! Hope this helps!
P.S. You won't be able to factor this one, which is why I suggested using this to find the roots instead.
A: Do you know about completing squares? Write your expression as $$11\left(x^2+\frac{13}{11}x\right)-7.$$ The goal is to write this as a difference of squares, which is easy to factorise. To this end, complete the square on the expression in brackets, which becomes $$x^2+2x\frac{13}{22}+\left(\frac{13}{22}\right)^2-\left(\frac{13}{22}\right)^2=\left(x+\frac{13}{22}\right)^2-\left(\frac{13}{22}\right)^2.$$
Thus our expression becomes $$11\left(x+\frac{13}{22}\right)^2-11\left(\frac{13}{22}\right)^2-7=\left(\sqrt {11}\left(x+\frac{13}{22}\right)\right)^2-\frac{169}{44}-7=\left(\sqrt {11}\left(x+\frac{13}{22}\right)\right)^2-\left(\sqrt{\frac{477}{44}}\right)^2,$$ from where you should now be able to continue.
