s^2+23s+22=0

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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{-44}{2}
=-22
$


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

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