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6f^2+13f-5=0
a = 6; b = 13; c = -5;
Δ = b2-4ac
Δ = 132-4·6·(-5)
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$$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-17}{2*6}=\frac{-30}{12} =-2+1/2 $$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+17}{2*6}=\frac{4}{12} =1/3 $
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