(3/5)(n+25)=10+(2/5)n

Solution for (3/5)(n+25)=10+(2/5)n equation:

(3/5)(n+25)=10+(2/5)n

We move all terms to the left:

(3/5)(n+25)-(10+(2/5)n)=0


Domain of the equation: 5)(n+25)!=0

n∈R



Domain of the equation: 5)n)!=0

n!=0/1

n!=0

n∈R


We add all the numbers together, and all the variables

(+3/5)(n+25)-(10+(+2/5)n)=0

We multiply parentheses ..

(+3n^2+3/5*25)-(10+(+2/5)n)=0

We calculate fractions


(3n^2+15n)/625n^2+()/625n^2=0

We multiply all the terms by the denominator


(3n^2+15n)+()=0

We add all the numbers together, and all the variables

(3n^2+15n)=0

We get rid of parentheses

3n^2+15n=0

a = 3; b = 15; c = 0;
Δ = b2-4ac
Δ = 152-4·3·0
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-15}{2*3}=\frac{-30}{6}
=-5
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+15}{2*3}=\frac{0}{6}
=0
$