-c^2+4c=-16

Solution for -c^2+4c=-16 equation:

-c^2+4c=-16

We move all terms to the left:

-c^2+4c-(-16)=0

We add all the numbers together, and all the variables

-1c^2+4c+16=0

a = -1; b = 4; c = +16;
Δ = b2-4ac
Δ = 42-4·(-1)·16
Δ = 80
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:

$\sqrt{\Delta}=\sqrt{80}=\sqrt{16*5}=\sqrt{16}*\sqrt{5}=4\sqrt{5}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{5}}{2*-1}=\frac{-4-4\sqrt{5}}{-2}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{5}}{2*-1}=\frac{-4+4\sqrt{5}}{-2}
$