(8n-5)(n+1)=0

Solution for (8n-5)(n+1)=0 equation:

(8n-5)(n+1)=0

We multiply parentheses ..

(+8n^2+8n-5n-5)=0

We get rid of parentheses

8n^2+8n-5n-5=0

We add all the numbers together, and all the variables

8n^2+3n-5=0

a = 8; b = 3; c = -5;
Δ = b2-4ac
Δ = 32-4·8·(-5)
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-13}{2*8}=\frac{-16}{16}
=-1
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+13}{2*8}=\frac{10}{16}
=5/8
$