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5f^2+24f-5=0
a = 5; b = 24; c = -5;
Δ = b2-4ac
Δ = 242-4·5·(-5)
Δ = 676
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{676}=26$$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-26}{2*5}=\frac{-50}{10} =-5 $$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+26}{2*5}=\frac{2}{10} =1/5 $
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