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r^2-5r-66=0
a = 1; b = -5; c = -66;
Δ = b2-4ac
Δ = -52-4·1·(-66)
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$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-17}{2*1}=\frac{-12}{2} =-6 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+17}{2*1}=\frac{22}{2} =11 $
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