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28r^2+17r-3=0
a = 28; b = 17; c = -3;
Δ = b2-4ac
Δ = 172-4·28·(-3)
Δ = 625
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{625}=25$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-25}{2*28}=\frac{-42}{56} =-3/4 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+25}{2*28}=\frac{8}{56} =1/7 $
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