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(18r^2)-33r+14=0
a = 18; b = -33; c = +14;
Δ = b2-4ac
Δ = -332-4·18·14
Δ = 81
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{81}=9$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-33)-9}{2*18}=\frac{24}{36} =2/3 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-33)+9}{2*18}=\frac{42}{36} =1+1/6 $
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