k^2+9k+16=0

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Solution for k^2+9k+16=0 equation: 



k^2+9k+16=0

a = 1; b = 9; c = +16;
Δ = b2-4ac
Δ = 92-4·1·16
Δ = 17
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-\sqrt{17}}{2*1}=\frac{-9-\sqrt{17}}{2}
$


$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+\sqrt{17}}{2*1}=\frac{-9+\sqrt{17}}{2}
$

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