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5r^2+28r+15=0
a = 5; b = 28; c = +15;
Δ = b2-4ac
Δ = 282-4·5·15
Δ = 484
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{484}=22$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-22}{2*5}=\frac{-50}{10} =-5 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+22}{2*5}=\frac{-6}{10} =-3/5 $
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