(2/5)n+4=20

Solution for (2/5)n+4=20 equation:

(2/5)n+4=20

We move all terms to the left:

(2/5)n+4-(20)=0


Domain of the equation: 5)n!=0

n!=0/1

n!=0

n∈R


We add all the numbers together, and all the variables

(+2/5)n+4-20=0

We add all the numbers together, and all the variables

(+2/5)n-16=0

We multiply parentheses

2n^2-16=0

a = 2; b = 0; c = -16;
Δ = b2-4ac
Δ = 02-4·2·(-16)
Δ = 128
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:

$\sqrt{\Delta}=\sqrt{128}=\sqrt{64*2}=\sqrt{64}*\sqrt{2}=8\sqrt{2}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{2}}{2*2}=\frac{0-8\sqrt{2}}{4}
=-\frac{8\sqrt{2}}{4}
=-2\sqrt{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{2}}{2*2}=\frac{0+8\sqrt{2}}{4}
=\frac{8\sqrt{2}}{4}
=2\sqrt{2}
$