n*2^n=160

Solution for n*2^n=160 equation:

n*2^n=160

We move all terms to the left:

n*2^n-(160)=0

Wy multiply elements

2n^2-160=0

a = 2; b = 0; c = -160;
Δ = b2-4ac
Δ = 02-4·2·(-160)
Δ = 1280
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:

$\sqrt{\Delta}=\sqrt{1280}=\sqrt{256*5}=\sqrt{256}*\sqrt{5}=16\sqrt{5}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{5}}{2*2}=\frac{0-16\sqrt{5}}{4}
=-\frac{16\sqrt{5}}{4}
=-4\sqrt{5}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{5}}{2*2}=\frac{0+16\sqrt{5}}{4}
=\frac{16\sqrt{5}}{4}
=4\sqrt{5}
$