(c-2)(7c-5)=0

Solution for (c-2)(7c-5)=0 equation:

(c-2)(7c-5)=0

We multiply parentheses ..

(+7c^2-5c-14c+10)=0

We get rid of parentheses

7c^2-5c-14c+10=0

We add all the numbers together, and all the variables

7c^2-19c+10=0

a = 7; b = -19; c = +10;
Δ = b2-4ac
Δ = -192-4·7·10
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-9}{2*7}=\frac{10}{14}
=5/7
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+9}{2*7}=\frac{28}{14}
=2
$