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r^2+13r-40=0
a = 1; b = 13; c = -40;
Δ = b2-4ac
Δ = 132-4·1·(-40)
Δ = 329
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}$$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-\sqrt{329}}{2*1}=\frac{-13-\sqrt{329}}{2} $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+\sqrt{329}}{2*1}=\frac{-13+\sqrt{329}}{2} $
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