s^2+14s+33=0

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Solution for s^2+14s+33=0 equation: 



s^2+14s+33=0

a = 1; b = 14; c = +33;
Δ = b2-4ac
Δ = 142-4·1·33
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$


$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-8}{2*1}=\frac{-22}{2}
=-11
$


$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+8}{2*1}=\frac{-6}{2}
=-3
$

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