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w^2+4w-285=0
a = 1; b = 4; c = -285;
Δ = b2-4ac
Δ = 42-4·1·(-285)
Δ = 1156
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{1156}=34$$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-34}{2*1}=\frac{-38}{2} =-19 $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+34}{2*1}=\frac{30}{2} =15 $
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