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3f^2+10f-13=0
a = 3; b = 10; c = -13;
Δ = b2-4ac
Δ = 102-4·3·(-13)
Δ = 256
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{256}=16$$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-16}{2*3}=\frac{-26}{6} =-4+1/3 $$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+16}{2*3}=\frac{6}{6} =1 $
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