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3r^2-23r-8=0
a = 3; b = -23; c = -8;
Δ = b2-4ac
Δ = -232-4·3·(-8)
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$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-23)-25}{2*3}=\frac{-2}{6} =-1/3 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23)+25}{2*3}=\frac{48}{6} =8 $
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