3c(c+9)=2c+12

Solution for 3c(c+9)=2c+12 equation:

3c(c+9)=2c+12

We move all terms to the left:

3c(c+9)-(2c+12)=0

We multiply parentheses

3c^2+27c-(2c+12)=0

We get rid of parentheses

3c^2+27c-2c-12=0

We add all the numbers together, and all the variables

3c^2+25c-12=0

a = 3; b = 25; c = -12;
Δ = b2-4ac
Δ = 252-4·3·(-12)
Δ = 769
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{-(25)-\sqrt{769}}{2*3}=\frac{-25-\sqrt{769}}{6}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+\sqrt{769}}{2*3}=\frac{-25+\sqrt{769}}{6}
$