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3f^2-34f+11=0
a = 3; b = -34; c = +11;
Δ = b2-4ac
Δ = -342-4·3·11
Δ = 1024
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{1024}=32$$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-34)-32}{2*3}=\frac{2}{6} =1/3 $$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-34)+32}{2*3}=\frac{66}{6} =11 $
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