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

Solution for 8(n+2)(n-7)=0 equation:

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

We multiply parentheses ..

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

We multiply parentheses

8n^2-56n+16n-112=0

We add all the numbers together, and all the variables

8n^2-40n-112=0

a = 8; b = -40; c = -112;
Δ = b2-4ac
Δ = -402-4·8·(-112)
Δ = 5184
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{5184}=72$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-72}{2*8}=\frac{-32}{16}
=-2
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+72}{2*8}=\frac{112}{16}
=7
$