(7n+4)(n-5)=0

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

(7n+4)(n-5)=0

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

7n^2-31n-20=0

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

$\sqrt{\Delta}=\sqrt{1521}=39$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-31)-39}{2*7}=\frac{-8}{14}
=-4/7
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-31)+39}{2*7}=\frac{70}{14}
=5
$