36=c^2+4c

Solution for 36=c^2+4c equation:

36=c^2+4c

We move all terms to the left:

36-(c^2+4c)=0

We get rid of parentheses

-c^2-4c+36=0

We add all the numbers together, and all the variables

-1c^2-4c+36=0

a = -1; b = -4; c = +36;
Δ = b2-4ac
Δ = -42-4·(-1)·36
Δ = 160
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{160}=\sqrt{16*10}=\sqrt{16}*\sqrt{10}=4\sqrt{10}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4\sqrt{10}}{2*-1}=\frac{4-4\sqrt{10}}{-2}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{10}}{2*-1}=\frac{4+4\sqrt{10}}{-2}
$