2c(c+3)+5c=c-12

Solution for 2c(c+3)+5c=c-12 equation:

2c(c+3)+5c=c-12

We move all terms to the left:

2c(c+3)+5c-(c-12)=0

We add all the numbers together, and all the variables

5c+2c(c+3)-(c-12)=0

We multiply parentheses

2c^2+5c+6c-(c-12)=0

We get rid of parentheses

2c^2+5c+6c-c+12=0

We add all the numbers together, and all the variables

2c^2+10c+12=0

a = 2; b = 10; c = +12;
Δ = b2-4ac
Δ = 102-4·2·12
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2}{2*2}=\frac{-12}{4}
=-3
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2}{2*2}=\frac{-8}{4}
=-2
$