40=n(2n+3)

Solution for 40=n(2n+3) equation:

40=n(2n+3)

We move all terms to the left:

40-(n(2n+3))=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^2-3n+40=0

a = -2; b = -3; c = +40;
Δ = b2-4ac
Δ = -32-4·(-2)·40
Δ = 329
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-\sqrt{329}}{2*-2}=\frac{3-\sqrt{329}}{-4}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{329}}{2*-2}=\frac{3+\sqrt{329}}{-4}
$