11a+a^2+18=0

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Solution for 11a+a^2+18=0 equation: 



11a+a^2+18=0

a = 1; b = 11; c = +18;
Δ = b2-4ac
Δ = 112-4·1·18
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$


$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-7}{2*1}=\frac{-18}{2}
=-9
$


$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+7}{2*1}=\frac{-4}{2}
=-2
$

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