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w^2+w-306=0
a = 1; b = 1; c = -306;
Δ = b2-4ac
Δ = 12-4·1·(-306)
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$$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-35}{2*1}=\frac{-36}{2} =-18 $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+35}{2*1}=\frac{34}{2} =17 $
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