9a+a^2+18=0

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Solution for 9a+a^2+18=0 equation:



9a+a^2+18=0
a = 1; b = 9; c = +18;
Δ = b2-4ac
Δ = 92-4·1·18
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-3}{2*1}=\frac{-12}{2} =-6 $
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+3}{2*1}=\frac{-6}{2} =-3 $

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