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3r^2-20r-7=0
a = 3; b = -20; c = -7;
Δ = b2-4ac
Δ = -202-4·3·(-7)
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$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-22}{2*3}=\frac{-2}{6} =-1/3 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+22}{2*3}=\frac{42}{6} =7 $
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