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3r^2-24r-315=0
a = 3; b = -24; c = -315;
Δ = b2-4ac
Δ = -242-4·3·(-315)
Δ = 4356
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{4356}=66$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-66}{2*3}=\frac{-42}{6} =-7 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+66}{2*3}=\frac{90}{6} =15 $
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