c^2+3c=3c+16

Solution for c^2+3c=3c+16 equation:

c^2+3c=3c+16

We move all terms to the left:

c^2+3c-(3c+16)=0

We get rid of parentheses

c^2+3c-3c-16=0

We add all the numbers together, and all the variables

c^2-16=0

a = 1; b = 0; c = -16;
Δ = b2-4ac
Δ = 02-4·1·(-16)
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8}{2*1}=\frac{-8}{2}
=-4
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8}{2*1}=\frac{8}{2}
=4
$