s^2-23s+22=0

Solution for s^2-23s+22=0 equation:

s^2-23s+22=0

a = 1; b = -23; c = +22;
Δ = b2-4ac
Δ = -232-4·1·22
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{441}=21$
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-23)-21}{2*1}=\frac{2}{2}
=1
$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23)+21}{2*1}=\frac{44}{2}
=22
$