8n(4n=5)=11

Solution for 8n(4n=5)=11 equation:

8n(4n=5)=11

We move all terms to the left:

8n(4n-(5))=0

We multiply parentheses

32n^2-40n=0

a = 32; b = -40; c = 0;
Δ = b2-4ac
Δ = -402-4·32·0
Δ = 1600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1600}=40$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-40}{2*32}=\frac{0}{64}
=0
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+40}{2*32}=\frac{80}{64}
=1+1/4
$