5(n2+n)=3(n2-2n)

Solution for 5(n2+n)=3(n2-2n) equation:

5(n2+n)=3(n2-2n)

We move all terms to the left:

5(n2+n)-(3(n2-2n))=0

We add all the numbers together, and all the variables

5(+n^2+n)-(3(+n^2-2n))=0

We multiply parentheses

5n^2-(3(+n^2-2n))+5n=0


We calculate terms in parentheses: -(3(+n^2-2n)), so:

3(+n^2-2n)

We multiply parentheses

3n^2-6n

Back to the equation:

-(3n^2-6n)


We add all the numbers together, and all the variables

5n^2+5n-(3n^2-6n)=0

We get rid of parentheses

5n^2-3n^2+5n+6n=0

We add all the numbers together, and all the variables

2n^2+11n=0

a = 2; b = 11; c = 0;
Δ = b2-4ac
Δ = 112-4·2·0
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-11}{2*2}=\frac{-22}{4}
=-5+1/2
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+11}{2*2}=\frac{0}{4}
=0
$