902=0.5n(n-3)

Solution for 902=0.5n(n-3) equation:

902=0.5n(n-3)

We move all terms to the left:

902-(0.5n(n-3))=0


We calculate terms in parentheses: -(0.5n(n-3)), so:

0.5n(n-3)

We multiply parentheses

0n^2+0n

We add all the numbers together, and all the variables

n^2+n

Back to the equation:

-(n^2+n)


We get rid of parentheses

-n^2-n+902=0

We add all the numbers together, and all the variables

-1n^2-1n+902=0

a = -1; b = -1; c = +902;
Δ = b2-4ac
Δ = -12-4·(-1)·902
Δ = 3609
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{3609}=\sqrt{9*401}=\sqrt{9}*\sqrt{401}=3\sqrt{401}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-3\sqrt{401}}{2*-1}=\frac{1-3\sqrt{401}}{-2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+3\sqrt{401}}{2*-1}=\frac{1+3\sqrt{401}}{-2}
$