b^2+18b+80=0

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Solution for b^2+18b+80=0 equation: 



b^2+18b+80=0

a = 1; b = 18; c = +80;
Δ = b2-4ac
Δ = 182-4·1·80
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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


$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-2}{2*1}=\frac{-20}{2}
=-10
$


$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+2}{2*1}=\frac{-16}{2}
=-8
$

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