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3r^2=12
We move all terms to the left:
3r^2-(12)=0
a = 3; b = 0; c = -12;
Δ = b2-4ac
Δ = 02-4·3·(-12)
Δ = 144
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{144}=12$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12}{2*3}=\frac{-12}{6} =-2 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12}{2*3}=\frac{12}{6} =2 $
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