(7n-4)(n+2)=0

Solution for (7n-4)(n+2)=0 equation:

(7n-4)(n+2)=0

We multiply parentheses ..

(+7n^2+14n-4n-8)=0

We get rid of parentheses

7n^2+14n-4n-8=0

We add all the numbers together, and all the variables

7n^2+10n-8=0

a = 7; b = 10; c = -8;
Δ = b2-4ac
Δ = 102-4·7·(-8)
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-18}{2*7}=\frac{-28}{14}
=-2
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+18}{2*7}=\frac{8}{14}
=4/7
$