3c+c(8-c)=24

Solution for 3c+c(8-c)=24 equation:

3c+c(8-c)=24

We move all terms to the left:

3c+c(8-c)-(24)=0

We add all the numbers together, and all the variables

3c+c(-1c+8)-24=0

We multiply parentheses

-1c^2+3c+8c-24=0

We add all the numbers together, and all the variables

-1c^2+11c-24=0

a = -1; b = 11; c = -24;
Δ = b2-4ac
Δ = 112-4·(-1)·(-24)
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-5}{2*-1}=\frac{-16}{-2}
=+8
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+5}{2*-1}=\frac{-6}{-2}
=+3
$