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18=2f^2
We move all terms to the left:
18-(2f^2)=0
a = -2; b = 0; c = +18;
Δ = b2-4ac
Δ = 02-4·(-2)·18
Δ = 144
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{144}=12$$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12}{2*-2}=\frac{-12}{-4} =+3 $$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12}{2*-2}=\frac{12}{-4} =-3 $
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