(n^2)-5n+4=0

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Solution for (n^2)-5n+4=0 equation: 



(n^2)-5n+4=0

a = 1; b = -5; c = +4;
Δ = b2-4ac
Δ = -52-4·1·4
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$


$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-3}{2*1}=\frac{2}{2}
=1
$


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

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