1n^2-4n-5=4n

Solution for 1n^2-4n-5=4n equation:

1n^2-4n-5=4n

We move all terms to the left:

1n^2-4n-5-(4n)=0

We add all the numbers together, and all the variables

n^2-8n-5=0

a = 1; b = -8; c = -5;
Δ = b2-4ac
Δ = -82-4·1·(-5)
Δ = 84
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:

$\sqrt{\Delta}=\sqrt{84}=\sqrt{4*21}=\sqrt{4}*\sqrt{21}=2\sqrt{21}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-2\sqrt{21}}{2*1}=\frac{8-2\sqrt{21}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+2\sqrt{21}}{2*1}=\frac{8+2\sqrt{21}}{2}
$