36b^2+12b-1=0

Solution for 36b^2+12b-1=0 equation:

36b^2+12b-1=0

a = 36; b = 12; c = -1;
Δ = b2-4ac
Δ = 122-4·36·(-1)
Δ = 288
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{288}=\sqrt{144*2}=\sqrt{144}*\sqrt{2}=12\sqrt{2}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-12\sqrt{2}}{2*36}=\frac{-12-12\sqrt{2}}{72}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+12\sqrt{2}}{2*36}=\frac{-12+12\sqrt{2}}{72}
$