w^2+19w+90=0

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Solution for w^2+19w+90=0 equation: 



w^2+19w+90=0

a = 1; b = 19; c = +90;
Δ = b2-4ac
Δ = 192-4·1·90
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$


$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-1}{2*1}=\frac{-20}{2}
=-10
$


$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+1}{2*1}=\frac{-18}{2}
=-9
$

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