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25f^2-25=0
a = 25; b = 0; c = -25;
Δ = b2-4ac
Δ = 02-4·25·(-25)
Δ = 2500
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{2500}=50$$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-50}{2*25}=\frac{-50}{50} =-1 $$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+50}{2*25}=\frac{50}{50} =1 $
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