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w^2+13w-264=0
a = 1; b = 13; c = -264;
Δ = b2-4ac
Δ = 132-4·1·(-264)
Δ = 1225
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{1225}=35$$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-35}{2*1}=\frac{-48}{2} =-24 $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+35}{2*1}=\frac{22}{2} =11 $
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