(4n+1)=n(2n+3)

Solution for (4n+1)=n(2n+3) equation:

(4n+1)=n(2n+3)

We move all terms to the left:

(4n+1)-(n(2n+3))=0

We get rid of parentheses

4n-(n(2n+3))+1=0


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

n(2n+3)

We multiply parentheses

2n^2+3n

Back to the equation:

-(2n^2+3n)


We get rid of parentheses

-2n^2+4n-3n+1=0

We add all the numbers together, and all the variables

-2n^2+n+1=0

a = -2; b = 1; c = +1;
Δ = b2-4ac
Δ = 12-4·(-2)·1
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-3}{2*-2}=\frac{-4}{-4}
=1
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+3}{2*-2}=\frac{2}{-4}
=-1/2
$