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300=3r^2
We move all terms to the left:
300-(3r^2)=0
a = -3; b = 0; c = +300;
Δ = b2-4ac
Δ = 02-4·(-3)·300
Δ = 3600
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{3600}=60$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-60}{2*-3}=\frac{-60}{-6} =+10 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+60}{2*-3}=\frac{60}{-6} =-10 $
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