6b^2+b=1

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

6b^2+b=1

We move all terms to the left:

6b^2+b-(1)=0

a = 6; b = 1; c = -1;
Δ = b2-4ac
Δ = 12-4·6·(-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{-(1)-5}{2*6}=\frac{-6}{12}
=-1/2
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+5}{2*6}=\frac{4}{12}
=1/3
$