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2f^2+33f+16=0
a = 2; b = 33; c = +16;
Δ = b2-4ac
Δ = 332-4·2·16
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$$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(33)-31}{2*2}=\frac{-64}{4} =-16 $$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(33)+31}{2*2}=\frac{-2}{4} =-1/2 $
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