(3c-7)(1c+1)=0

Solution for (3c-7)(1c+1)=0 equation:

(3c-7)(1c+1)=0

We add all the numbers together, and all the variables

(3c-7)(c+1)=0

We multiply parentheses ..

(+3c^2+3c-7c-7)=0

We get rid of parentheses

3c^2+3c-7c-7=0

We add all the numbers together, and all the variables

3c^2-4c-7=0

a = 3; b = -4; c = -7;
Δ = b2-4ac
Δ = -42-4·3·(-7)
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-10}{2*3}=\frac{-6}{6}
=-1
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+10}{2*3}=\frac{14}{6}
=2+1/3
$