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

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

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

We multiply parentheses ..

(+3c^2+9c-5c-15)=0

We get rid of parentheses

3c^2+9c-5c-15=0

We add all the numbers together, and all the variables

3c^2+4c-15=0

a = 3; b = 4; c = -15;
Δ = b2-4ac
Δ = 42-4·3·(-15)
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-14}{2*3}=\frac{-18}{6}
=-3
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+14}{2*3}=\frac{10}{6}
=1+2/3
$