c^2+4-(c+4)=32

Solution for c^2+4-(c+4)=32 equation:

c^2+4-(c+4)=32

We move all terms to the left:

c^2+4-(c+4)-(32)=0

We add all the numbers together, and all the variables

c^2-(c+4)-28=0

We get rid of parentheses

c^2-c-4-28=0

We add all the numbers together, and all the variables

c^2-1c-32=0

a = 1; b = -1; c = -32;
Δ = b2-4ac
Δ = -12-4·1·(-32)
Δ = 129
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-\sqrt{129}}{2*1}=\frac{1-\sqrt{129}}{2}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{129}}{2*1}=\frac{1+\sqrt{129}}{2}
$