-7/2(2n-3)+4n=3/2(5+2n)

Solution for -7/2(2n-3)+4n=3/2(5+2n) equation:

-7/2(2n-3)+4n=3/2(5+2n)

We move all terms to the left:

-7/2(2n-3)+4n-(3/2(5+2n))=0


Domain of the equation: 2(2n-3)!=0

n∈R



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

n∈R


We add all the numbers together, and all the variables

-7/2(2n-3)+4n-(3/2(2n+5))=0

We add all the numbers together, and all the variables

4n-7/2(2n-3)-(3/2(2n+5))=0

We calculate fractions


4n+(-14n2/(2(2n-3)*2(2n+5)))+(-6n2/(2(2n-3)*2(2n+5)))=0


We calculate terms in parentheses: +(-14n2/(2(2n-3)*2(2n+5))), so:

-14n2/(2(2n-3)*2(2n+5))

We multiply all the terms by the denominator


-14n2

We add all the numbers together, and all the variables

-14n^2

Back to the equation:

+(-14n^2)



We calculate terms in parentheses: +(-6n2/(2(2n-3)*2(2n+5))), so:

-6n2/(2(2n-3)*2(2n+5))

We multiply all the terms by the denominator


-6n2

We add all the numbers together, and all the variables

-6n^2

Back to the equation:

+(-6n^2)


We get rid of parentheses

-14n^2-6n^2+4n=0

We add all the numbers together, and all the variables

-20n^2+4n=0

a = -20; b = 4; c = 0;
Δ = b2-4ac
Δ = 42-4·(-20)·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4}{2*-20}=\frac{-8}{-40}
=1/5
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4}{2*-20}=\frac{0}{-40}
=0
$