8n(3n+4)=11

Solution for 8n(3n+4)=11 equation:

8n(3n+4)=11

We move all terms to the left:

8n(3n+4)-(11)=0

We multiply parentheses

24n^2+32n-11=0

a = 24; b = 32; c = -11;
Δ = b2-4ac
Δ = 322-4·24·(-11)
Δ = 2080
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:

$\sqrt{\Delta}=\sqrt{2080}=\sqrt{16*130}=\sqrt{16}*\sqrt{130}=4\sqrt{130}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-4\sqrt{130}}{2*24}=\frac{-32-4\sqrt{130}}{48}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+4\sqrt{130}}{2*24}=\frac{-32+4\sqrt{130}}{48}
$