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(6r)^2+r-40=0
a = 6; b = 1; c = -40;
Δ = b2-4ac
Δ = 12-4·6·(-40)
Δ = 961
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{961}=31$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-31}{2*6}=\frac{-32}{12} =-2+2/3 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+31}{2*6}=\frac{30}{12} =2+1/2 $
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