7/10n+3/2=3/5n+12

Solution for 7/10n+3/2=3/5n+12 equation:

7/10n+3/2=3/5n+12

We move all terms to the left:

7/10n+3/2-(3/5n+12)=0


Domain of the equation: 10n!=0

n!=0/10

n!=0

n∈R



Domain of the equation: 5n+12)!=0

n∈R


We get rid of parentheses

7/10n-3/5n-12+3/2=0

We calculate fractions


750n^2/200n^2+140n/200n^2+(-120n)/200n^2-12=0

We multiply all the terms by the denominator


750n^2+140n+(-120n)-12*200n^2=0

Wy multiply elements

750n^2-2400n^2+140n+(-120n)=0

We get rid of parentheses

750n^2-2400n^2+140n-120n=0

We add all the numbers together, and all the variables

-1650n^2+20n=0

a = -1650; b = 20; c = 0;
Δ = b2-4ac
Δ = 202-4·(-1650)·0
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-20}{2*-1650}=\frac{-40}{-3300}
=2/165
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+20}{2*-1650}=\frac{0}{-3300}
=0
$