k^2+20k+99=0

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Solution for k^2+20k+99=0 equation: 



k^2+20k+99=0

a = 1; b = 20; c = +99;
Δ = b2-4ac
Δ = 202-4·1·99
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$


$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-2}{2*1}=\frac{-22}{2}
=-11
$


$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+2}{2*1}=\frac{-18}{2}
=-9
$

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