3c=2c(c-1)

Solution for 3c=2c(c-1) equation:

3c=2c(c-1)

We move all terms to the left:

3c-(2c(c-1))=0


We calculate terms in parentheses: -(2c(c-1)), so:

2c(c-1)

We multiply parentheses

2c^2-2c

Back to the equation:

-(2c^2-2c)


We get rid of parentheses

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

We add all the numbers together, and all the variables

-2c^2+5c=0

a = -2; b = 5; c = 0;
Δ = b2-4ac
Δ = 52-4·(-2)·0
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-5}{2*-2}=\frac{-10}{-4}
=2+1/2
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+5}{2*-2}=\frac{0}{-4}
=0
$