(6c)(2)=2c+40

Solution for (6c)(2)=2c+40 equation:

(6c)(2)=2c+40

We move all terms to the left:

(6c)(2)-(2c+40)=0

We add all the numbers together, and all the variables

6c^2-(2c+40)=0

We get rid of parentheses

6c^2-2c-40=0

a = 6; b = -2; c = -40;
Δ = b2-4ac
Δ = -22-4·6·(-40)
Δ = 964
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{964}=\sqrt{4*241}=\sqrt{4}*\sqrt{241}=2\sqrt{241}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{241}}{2*6}=\frac{2-2\sqrt{241}}{12}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{241}}{2*6}=\frac{2+2\sqrt{241}}{12}
$