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)-(11)=0

We multiply parentheses

32n^2+40n-11=0

a = 32; b = 40; c = -11;
Δ = b2-4ac
Δ = 402-4·32·(-11)
Δ = 3008
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{3008}=\sqrt{64*47}=\sqrt{64}*\sqrt{47}=8\sqrt{47}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-8\sqrt{47}}{2*32}=\frac{-40-8\sqrt{47}}{64}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+8\sqrt{47}}{2*32}=\frac{-40+8\sqrt{47}}{64}
$