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3r^2=363
We move all terms to the left:
3r^2-(363)=0
a = 3; b = 0; c = -363;
Δ = b2-4ac
Δ = 02-4·3·(-363)
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$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-66}{2*3}=\frac{-66}{6} =-11 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+66}{2*3}=\frac{66}{6} =11 $
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