(c+3)(3c+7)=0

Solution for (c+3)(3c+7)=0 equation:

(c+3)(3c+7)=0

We multiply parentheses ..

(+3c^2+7c+9c+21)=0

We get rid of parentheses

3c^2+7c+9c+21=0

We add all the numbers together, and all the variables

3c^2+16c+21=0

a = 3; b = 16; c = +21;
Δ = b2-4ac
Δ = 162-4·3·21
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{-(16)-2}{2*3}=\frac{-18}{6}
=-3
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+2}{2*3}=\frac{-14}{6}
=-2+1/3
$