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w^2+10w-336=0
a = 1; b = 10; c = -336;
Δ = b2-4ac
Δ = 102-4·1·(-336)
Δ = 1444
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{1444}=38$$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-38}{2*1}=\frac{-48}{2} =-24 $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+38}{2*1}=\frac{28}{2} =14 $
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