s^2-3s+1=0

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Solution for s^2-3s+1=0 equation: 



s^2-3s+1=0

a = 1; b = -3; c = +1;
Δ = b2-4ac
Δ = -32-4·1·1
Δ = 5
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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


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

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