3c-2c(c-9)=2+2c

Solution for 3c-2c(c-9)=2+2c equation:

3c-2c(c-9)=2+2c

We move all terms to the left:

3c-2c(c-9)-(2+2c)=0

We add all the numbers together, and all the variables

3c-2c(c-9)-(2c+2)=0

We multiply parentheses

-2c^2+3c+18c-(2c+2)=0

We get rid of parentheses

-2c^2+3c+18c-2c-2=0

We add all the numbers together, and all the variables

-2c^2+19c-2=0

a = -2; b = 19; c = -2;
Δ = b2-4ac
Δ = 192-4·(-2)·(-2)
Δ = 345
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{-(19)-\sqrt{345}}{2*-2}=\frac{-19-\sqrt{345}}{-4}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+\sqrt{345}}{2*-2}=\frac{-19+\sqrt{345}}{-4}
$