2c-3c(c-4)=c+2

Solution for 2c-3c(c-4)=c+2 equation:

2c-3c(c-4)=c+2

We move all terms to the left:

2c-3c(c-4)-(c+2)=0

We multiply parentheses

-3c^2+2c+12c-(c+2)=0

We get rid of parentheses

-3c^2+2c+12c-c-2=0

We add all the numbers together, and all the variables

-3c^2+13c-2=0

a = -3; b = 13; c = -2;
Δ = b2-4ac
Δ = 132-4·(-3)·(-2)
Δ = 145
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{-(13)-\sqrt{145}}{2*-3}=\frac{-13-\sqrt{145}}{-6}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+\sqrt{145}}{2*-3}=\frac{-13+\sqrt{145}}{-6}
$