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16f^2-11f=0
a = 16; b = -11; c = 0;
Δ = b2-4ac
Δ = -112-4·16·0
Δ = 121
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{121}=11$$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-11}{2*16}=\frac{0}{32} =0 $$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+11}{2*16}=\frac{22}{32} =11/16 $
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