4(2n-)n=5

Solution for 4(2n-)n=5 equation:

4(2n-)n=5

We move all terms to the left:

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

We add all the numbers together, and all the variables

4(+2n)n-5=0

We multiply parentheses

8n^2-5=0

a = 8; b = 0; c = -5;
Δ = b2-4ac
Δ = 02-4·8·(-5)
Δ = 160
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:

$\sqrt{\Delta}=\sqrt{160}=\sqrt{16*10}=\sqrt{16}*\sqrt{10}=4\sqrt{10}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{10}}{2*8}=\frac{0-4\sqrt{10}}{16}
=-\frac{4\sqrt{10}}{16}
=-\frac{\sqrt{10}}{4}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{10}}{2*8}=\frac{0+4\sqrt{10}}{16}
=\frac{4\sqrt{10}}{16}
=\frac{\sqrt{10}}{4}
$