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6r^2+r-12=0
a = 6; b = 1; c = -12;
Δ = b2-4ac
Δ = 12-4·6·(-12)
Δ = 289
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{289}=17$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-17}{2*6}=\frac{-18}{12} =-1+1/2 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+17}{2*6}=\frac{16}{12} =1+1/3 $
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