(4n-3)(n-8)=0

Solution for (4n-3)(n-8)=0 equation:

(4n-3)(n-8)=0

We multiply parentheses ..

(+4n^2-32n-3n+24)=0

We get rid of parentheses

4n^2-32n-3n+24=0

We add all the numbers together, and all the variables

4n^2-35n+24=0

a = 4; b = -35; c = +24;
Δ = b2-4ac
Δ = -352-4·4·24
Δ = 841
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{841}=29$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-35)-29}{2*4}=\frac{6}{8}
=3/4
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-35)+29}{2*4}=\frac{64}{8}
=8
$