8(c-9)(2c-12)-4c=6

Solution for 8(c-9)(2c-12)-4c=6 equation:

8(c-9)(2c-12)-4c=6

We move all terms to the left:

8(c-9)(2c-12)-4c-(6)=0

We add all the numbers together, and all the variables

-4c+8(c-9)(2c-12)-6=0

We multiply parentheses ..

8(+2c^2-12c-18c+108)-4c-6=0

We multiply parentheses

16c^2-96c-144c-4c+864-6=0

We add all the numbers together, and all the variables

16c^2-244c+858=0

a = 16; b = -244; c = +858;
Δ = b2-4ac
Δ = -2442-4·16·858
Δ = 4624
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4624}=68$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-244)-68}{2*16}=\frac{176}{32}
=5+1/2
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-244)+68}{2*16}=\frac{312}{32}
=9+3/4
$