2c(5-c)+12c=10c

Solution for 2c(5-c)+12c=10c equation:

2c(5-c)+12c=10c

We move all terms to the left:

2c(5-c)+12c-(10c)=0

We add all the numbers together, and all the variables

2c(-1c+5)+12c-10c=0

We add all the numbers together, and all the variables

2c+2c(-1c+5)=0

We multiply parentheses

-2c^2+2c+10c=0

We add all the numbers together, and all the variables

-2c^2+12c=0

a = -2; b = 12; c = 0;
Δ = b2-4ac
Δ = 122-4·(-2)·0
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-12}{2*-2}=\frac{-24}{-4}
=+6
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+12}{2*-2}=\frac{0}{-4}
=0
$