22s^2+9s-1=0

Solution for 22s^2+9s-1=0 equation:

22s^2+9s-1=0

a = 22; b = 9; c = -1;
Δ = b2-4ac
Δ = 92-4·22·(-1)
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-13}{2*22}=\frac{-22}{44}
=-1/2
$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+13}{2*22}=\frac{4}{44}
=1/11
$