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25f^2-46f=0
a = 25; b = -46; c = 0;
Δ = b2-4ac
Δ = -462-4·25·0
Δ = 2116
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{2116}=46$$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-46)-46}{2*25}=\frac{0}{50} =0 $$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-46)+46}{2*25}=\frac{92}{50} =1+21/25 $
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