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45r^2+65r+20=0
a = 45; b = 65; c = +20;
Δ = b2-4ac
Δ = 652-4·45·20
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$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(65)-25}{2*45}=\frac{-90}{90} =-1 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(65)+25}{2*45}=\frac{-40}{90} =-4/9 $
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