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48f^2+31f=0
a = 48; b = 31; c = 0;
Δ = b2-4ac
Δ = 312-4·48·0
Δ = 961
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{961}=31$$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(31)-31}{2*48}=\frac{-62}{96} =-31/48 $$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(31)+31}{2*48}=\frac{0}{96} =0 $
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