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3w^2+19w+20=0
a = 3; b = 19; c = +20;
Δ = b2-4ac
Δ = 192-4·3·20
Δ = 121
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{121}=11$$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-11}{2*3}=\frac{-30}{6} =-5 $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+11}{2*3}=\frac{-8}{6} =-1+1/3 $
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