c-4=c^2-16

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

c-4=c^2-16

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-1c^2+c+12=0

a = -1; b = 1; c = +12;
Δ = b2-4ac
Δ = 12-4·(-1)·12
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-7}{2*-1}=\frac{-8}{-2}
=+4
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+7}{2*-1}=\frac{6}{-2}
=-3
$