7/10n+3/5=2/5n+2

Solution for 7/10n+3/5=2/5n+2 equation:

7/10n+3/5=2/5n+2

We move all terms to the left:

7/10n+3/5-(2/5n+2)=0


Domain of the equation: 10n!=0

n!=0/10

n!=0

n∈R



Domain of the equation: 5n+2)!=0

n∈R


We get rid of parentheses

7/10n-2/5n-2+3/5=0

We calculate fractions


875n/1250n^2+(-20n)/1250n^2+30n/1250n^2-2=0

We multiply all the terms by the denominator


875n+(-20n)+30n-2*1250n^2=0

We add all the numbers together, and all the variables

905n+(-20n)-2*1250n^2=0

Wy multiply elements

-2500n^2+905n+(-20n)=0

We get rid of parentheses

-2500n^2+905n-20n=0

We add all the numbers together, and all the variables

-2500n^2+885n=0

a = -2500; b = 885; c = 0;
Δ = b2-4ac
Δ = 8852-4·(-2500)·0
Δ = 783225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{783225}=885$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(885)-885}{2*-2500}=\frac{-1770}{-5000}
=177/500
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(885)+885}{2*-2500}=\frac{0}{-5000}
=0
$