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6r^2+6r-120=0
a = 6; b = 6; c = -120;
Δ = b2-4ac
Δ = 62-4·6·(-120)
Δ = 2916
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{2916}=54$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-54}{2*6}=\frac{-60}{12} =-5 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+54}{2*6}=\frac{48}{12} =4 $
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