(-3c-7)(-3c-5)=0

Solution for (-3c-7)(-3c-5)=0 equation:

(-3c-7)(-3c-5)=0

We multiply parentheses ..

(+9c^2+15c+21c+35)=0

We get rid of parentheses

9c^2+15c+21c+35=0

We add all the numbers together, and all the variables

9c^2+36c+35=0

a = 9; b = 36; c = +35;
Δ = b2-4ac
Δ = 362-4·9·35
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(36)-6}{2*9}=\frac{-42}{18}
=-2+1/3
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+6}{2*9}=\frac{-30}{18}
=-1+2/3
$