56b2+b=1

Solution for 56b2+b=1 equation:

56b^2+b=1

We move all terms to the left:

56b^2+b-(1)=0

a = 56; b = 1; c = -1;
Δ = b2-4ac
Δ = 12-4·56·(-1)
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-15}{2*56}=\frac{-16}{112}
=-1/7
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+15}{2*56}=\frac{14}{112}
=1/8
$