-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}
$