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2w^2+13w+11=0
a = 2; b = 13; c = +11;
Δ = b2-4ac
Δ = 132-4·2·11
Δ = 81
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{81}=9$$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-9}{2*2}=\frac{-22}{4} =-5+1/2 $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+9}{2*2}=\frac{-4}{4} =-1 $
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