3b2-5b+1=0

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Solution for 3b2-5b+1=0 equation: 



3b^2-5b+1=0

a = 3; b = -5; c = +1;
Δ = b2-4ac
Δ = -52-4·3·1
Δ = 13
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-\sqrt{13}}{2*3}=\frac{5-\sqrt{13}}{6}
$


$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{13}}{2*3}=\frac{5+\sqrt{13}}{6}
$

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