C(c+2)-c(c-6)=10c

Solution for C(c+2)-c(c-6)=10c equation:

(C+2)-C(C-6)=10C

We move all terms to the left:

(C+2)-C(C-6)-(10C)=0

We add all the numbers together, and all the variables

-10C+(C+2)-C(C-6)=0

We multiply parentheses

-C^2-10C+(C+2)+6C=0

We get rid of parentheses

-C^2-10C+C+6C+2=0

We add all the numbers together, and all the variables

-1C^2-3C+2=0

a = -1; b = -3; c = +2;
Δ = b2-4ac
Δ = -32-4·(-1)·2
Δ = 17
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{-(-3)-\sqrt{17}}{2*-1}=\frac{3-\sqrt{17}}{-2}
$
$C_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{17}}{2*-1}=\frac{3+\sqrt{17}}{-2}
$