4w^2+19w+18=0

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Solution for 4w^2+19w+18=0 equation: 



4w^2+19w+18=0

a = 4; b = 19; c = +18;
Δ = b2-4ac
Δ = 192-4·4·18
Δ = 73
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-\sqrt{73}}{2*4}=\frac{-19-\sqrt{73}}{8}
$


$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+\sqrt{73}}{2*4}=\frac{-19+\sqrt{73}}{8}
$

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