2n+(1/2)n=10

Solution for 2n+(1/2)n=10 equation:

2n+(1/2)n=10

We move all terms to the left:

2n+(1/2)n-(10)=0


Domain of the equation: 2)n!=0

n!=0/1

n!=0

n∈R


We add all the numbers together, and all the variables

2n+(+1/2)n-10=0

We multiply parentheses

n^2+2n-10=0

a = 1; b = 2; c = -10;
Δ = b2-4ac
Δ = 22-4·1·(-10)
Δ = 44
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:

$\sqrt{\Delta}=\sqrt{44}=\sqrt{4*11}=\sqrt{4}*\sqrt{11}=2\sqrt{11}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{11}}{2*1}=\frac{-2-2\sqrt{11}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{11}}{2*1}=\frac{-2+2\sqrt{11}}{2}
$