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f^2-13f-300=0
a = 1; b = -13; c = -300;
Δ = b2-4ac
Δ = -132-4·1·(-300)
Δ = 1369
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{1369}=37$$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-37}{2*1}=\frac{-24}{2} =-12 $$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+37}{2*1}=\frac{50}{2} =25 $
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