-11=m-4,m^2+m

Solution for -11=m-4,m^2+m equation:

-11=m-4.m^2+m

We move all terms to the left:

-11-(m-4.m^2+m)=0

We get rid of parentheses

4.m^2-m-m-11=0

We add all the numbers together, and all the variables

4.m^2-2m-11=0

a = 4.; b = -2; c = -11;
Δ = b2-4ac
Δ = -22-4·4.·(-11)
Δ = 180
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{180}=\sqrt{36*5}=\sqrt{36}*\sqrt{5}=6\sqrt{5}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-6\sqrt{5}}{2*4.}=\frac{2-6\sqrt{5}}{8}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+6\sqrt{5}}{2*4.}=\frac{2+6\sqrt{5}}{8}
$