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3r^2-24=21r
We move all terms to the left:
3r^2-24-(21r)=0
a = 3; b = -21; c = -24;
Δ = b2-4ac
Δ = -212-4·3·(-24)
Δ = 729
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{729}=27$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-21)-27}{2*3}=\frac{-6}{6} =-1 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-21)+27}{2*3}=\frac{48}{6} =8 $
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