25b^2-20b-1=0

Solution for 25b^2-20b-1=0 equation:

25b^2-20b-1=0

a = 25; b = -20; c = -1;
Δ = b2-4ac
Δ = -202-4·25·(-1)
Δ = 500
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:

$\sqrt{\Delta}=\sqrt{500}=\sqrt{100*5}=\sqrt{100}*\sqrt{5}=10\sqrt{5}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-10\sqrt{5}}{2*25}=\frac{20-10\sqrt{5}}{50}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+10\sqrt{5}}{2*25}=\frac{20+10\sqrt{5}}{50}
$