902=0.5n(n-3)

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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} $

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