(5c+1)(4c-7)=0

Solution for (5c+1)(4c-7)=0 equation:

(5c+1)(4c-7)=0

We multiply parentheses ..

(+20c^2-35c+4c-7)=0

We get rid of parentheses

20c^2-35c+4c-7=0

We add all the numbers together, and all the variables

20c^2-31c-7=0

a = 20; b = -31; c = -7;
Δ = b2-4ac
Δ = -312-4·20·(-7)
Δ = 1521
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1521}=39$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-31)-39}{2*20}=\frac{-8}{40}
=-1/5
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-31)+39}{2*20}=\frac{70}{40}
=1+3/4
$