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r^2-23r-50=0
a = 1; b = -23; c = -50;
Δ = b2-4ac
Δ = -232-4·1·(-50)
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$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-23)-27}{2*1}=\frac{-4}{2} =-2 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23)+27}{2*1}=\frac{50}{2} =25 $
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