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25f^2+26f=0
a = 25; b = 26; c = 0;
Δ = b2-4ac
Δ = 262-4·25·0
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$$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(26)-26}{2*25}=\frac{-52}{50} =-1+1/25 $$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(26)+26}{2*25}=\frac{0}{50} =0 $
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