6b^2-5b+1=0

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Solution for 6b^2-5b+1=0 equation: 



6b^2-5b+1=0

a = 6; b = -5; c = +1;
Δ = b2-4ac
Δ = -52-4·6·1
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$


$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-1}{2*6}=\frac{4}{12}
=1/3
$


$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+1}{2*6}=\frac{6}{12}
=1/2
$

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