3b^2-13b+6=0

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Solution for 3b^2-13b+6=0 equation: 



3b^2-13b+6=0

a = 3; b = -13; c = +6;
Δ = b2-4ac
Δ = -132-4·3·6
Δ = 97
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{-(-13)-\sqrt{97}}{2*3}=\frac{13-\sqrt{97}}{6}
$


$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+\sqrt{97}}{2*3}=\frac{13+\sqrt{97}}{6}
$

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