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3r^2-12r=63
We move all terms to the left:
3r^2-12r-(63)=0
a = 3; b = -12; c = -63;
Δ = b2-4ac
Δ = -122-4·3·(-63)
Δ = 900
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{900}=30$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-30}{2*3}=\frac{-18}{6} =-3 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+30}{2*3}=\frac{42}{6} =7 $
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