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6f^2+20f-896=0
a = 6; b = 20; c = -896;
Δ = b2-4ac
Δ = 202-4·6·(-896)
Δ = 21904
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{21904}=148$$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-148}{2*6}=\frac{-168}{12} =-14 $$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+148}{2*6}=\frac{128}{12} =10+2/3 $
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