3b2+b-2=0

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Solution for 3b2+b-2=0 equation: 



3b^2+b-2=0

a = 3; b = 1; c = -2;
Δ = b2-4ac
Δ = 12-4·3·(-2)
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$


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


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

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