2w^2+15w+14=0

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Solution for 2w^2+15w+14=0 equation:



2w^2+15w+14=0
a = 2; b = 15; c = +14;
Δ = b2-4ac
Δ = 152-4·2·14
Δ = 113
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{-(15)-\sqrt{113}}{2*2}=\frac{-15-\sqrt{113}}{4} $
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+\sqrt{113}}{2*2}=\frac{-15+\sqrt{113}}{4} $

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