5c-4c-4(c-3c);c=6

Solution for 5c-4c-4(c-3c);c=6 equation:

5c-4c-4(c-3c)c=6

We move all terms to the left:

5c-4c-4(c-3c)c-(6)=0

We add all the numbers together, and all the variables

5c-4c-4(-2c)c-6=0

We add all the numbers together, and all the variables

c-4(-2c)c-6=0

We multiply parentheses

8c^2+c-6=0

a = 8; b = 1; c = -6;
Δ = b2-4ac
Δ = 12-4·8·(-6)
Δ = 193
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{193}}{2*8}=\frac{-1-\sqrt{193}}{16}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{193}}{2*8}=\frac{-1+\sqrt{193}}{16}
$