2w^2+18w+40=0

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Solution for 2w^2+18w+40=0 equation: 



2w^2+18w+40=0

a = 2; b = 18; c = +40;
Δ = b2-4ac
Δ = 182-4·2·40
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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


$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-2}{2*2}=\frac{-20}{4}
=-5
$


$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+2}{2*2}=\frac{-16}{4}
=-4
$

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