(n-5)(n-3)=0

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Solution for (n-5)(n-3)=0 equation:



(n-5)(n-3)=0
We multiply parentheses ..
(+n^2-3n-5n+15)=0
We get rid of parentheses
n^2-3n-5n+15=0
We add all the numbers together, and all the variables
n^2-8n+15=0
a = 1; b = -8; c = +15;
Δ = b2-4ac
Δ = -82-4·1·15
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-2}{2*1}=\frac{6}{2} =3 $
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+2}{2*1}=\frac{10}{2} =5 $

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