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10r^2-18r-4=0
a = 10; b = -18; c = -4;
Δ = b2-4ac
Δ = -182-4·10·(-4)
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$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-22}{2*10}=\frac{-4}{20} =-1/5 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+22}{2*10}=\frac{40}{20} =2 $
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