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5r^2+33r+18=0
a = 5; b = 33; c = +18;
Δ = b2-4ac
Δ = 332-4·5·18
Δ = 729
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{729}=27$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(33)-27}{2*5}=\frac{-60}{10} =-6 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(33)+27}{2*5}=\frac{-6}{10} =-3/5 $
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