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5w^2+15w-20=0
a = 5; b = 15; c = -20;
Δ = b2-4ac
Δ = 152-4·5·(-20)
Δ = 625
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{625}=25$$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-25}{2*5}=\frac{-40}{10} =-4 $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+25}{2*5}=\frac{10}{10} =1 $
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