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3w^2+25w+28=0
a = 3; b = 25; c = +28;
Δ = b2-4ac
Δ = 252-4·3·28
Δ = 289
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{289}=17$$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-17}{2*3}=\frac{-42}{6} =-7 $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+17}{2*3}=\frac{-8}{6} =-1+1/3 $
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