N(t)=250(2^(-t/10))

Solution for N(t)=250(2^(-t/10)) equation:

(N)=250(2^(-N/10))

We move all terms to the left:

(N)-(250(2^(-N/10)))=0

We multiply all the terms by the denominator


N*10)))-(250(2^(-N=0

We add all the numbers together, and all the variables

-1N+N*10)))-(250(2^(=0

Wy multiply elements

10N^2-1N=0

a = 10; b = -1; c = 0;
Δ = b2-4ac
Δ = -12-4·10·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$N_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$N_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$N_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-1}{2*10}=\frac{0}{20}
=0
$
$N_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+1}{2*10}=\frac{2}{20}
=1/10
$