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24r^2+19r+2=0
a = 24; b = 19; c = +2;
Δ = b2-4ac
Δ = 192-4·24·2
Δ = 169
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{169}=13$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-13}{2*24}=\frac{-32}{48} =-2/3 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+13}{2*24}=\frac{-6}{48} =-1/8 $
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