208=n(n-3)

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Solution for 208=n(n-3) equation:



208=n(n-3)
We move all terms to the left:
208-(n(n-3))=0
We calculate terms in parentheses: -(n(n-3)), so:
n(n-3)
We multiply parentheses
n^2-3n
Back to the equation:
-(n^2-3n)
We get rid of parentheses
-n^2+3n+208=0
We add all the numbers together, and all the variables
-1n^2+3n+208=0
a = -1; b = 3; c = +208;
Δ = b2-4ac
Δ = 32-4·(-1)·208
Δ = 841
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{841}=29$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-29}{2*-1}=\frac{-32}{-2} =+16 $
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+29}{2*-1}=\frac{26}{-2} =-13 $

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