6b^2+7b+1=0

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Solution for 6b^2+7b+1=0 equation: 



6b^2+7b+1=0

a = 6; b = 7; c = +1;
Δ = b2-4ac
Δ = 72-4·6·1
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{-(7)-5}{2*6}=\frac{-12}{12}
=-1
$


$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+5}{2*6}=\frac{-2}{12}
=-1/6
$

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