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2f^2+13f+4=0
a = 2; b = 13; c = +4;
Δ = b2-4ac
Δ = 132-4·2·4
Δ = 137
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}$$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-\sqrt{137}}{2*2}=\frac{-13-\sqrt{137}}{4} $$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+\sqrt{137}}{2*2}=\frac{-13+\sqrt{137}}{4} $
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