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f^2-19f+18=0
a = 1; b = -19; c = +18;
Δ = b2-4ac
Δ = -192-4·1·18
Δ = 289
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{289}=17$$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-17}{2*1}=\frac{2}{2} =1 $$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+17}{2*1}=\frac{36}{2} =18 $
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