8n(4n+5)=27

Solution for 8n(4n+5)=27 equation:

8n(4n+5)=27

We move all terms to the left:

8n(4n+5)-(27)=0

We multiply parentheses

32n^2+40n-27=0

a = 32; b = 40; c = -27;
Δ = b2-4ac
Δ = 402-4·32·(-27)
Δ = 5056
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:

$\sqrt{\Delta}=\sqrt{5056}=\sqrt{64*79}=\sqrt{64}*\sqrt{79}=8\sqrt{79}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-8\sqrt{79}}{2*32}=\frac{-40-8\sqrt{79}}{64}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+8\sqrt{79}}{2*32}=\frac{-40+8\sqrt{79}}{64}
$