(2c-5)(7c+1)=0

Solution for (2c-5)(7c+1)=0 equation:

(2c-5)(7c+1)=0

We multiply parentheses ..

(+14c^2+2c-35c-5)=0

We get rid of parentheses

14c^2+2c-35c-5=0

We add all the numbers together, and all the variables

14c^2-33c-5=0

a = 14; b = -33; c = -5;
Δ = b2-4ac
Δ = -332-4·14·(-5)
Δ = 1369
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1369}=37$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-33)-37}{2*14}=\frac{-4}{28}
=-1/7
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-33)+37}{2*14}=\frac{70}{28}
=2+1/2
$