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f(2)=3(1.09)8
We move all terms to the left:
f(2)-(3(1.09)8)=0
We add all the numbers together, and all the variables
f^2-26.16=0
a = 1; b = 0; c = -26.16;
Δ = b2-4ac
Δ = 02-4·1·(-26.16)
Δ = 104.64
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}$$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-\sqrt{104.64}}{2*1}=\frac{0-\sqrt{104.64}}{2} =-\frac{\sqrt{}}{2} $$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+\sqrt{104.64}}{2*1}=\frac{0+\sqrt{104.64}}{2} =\frac{\sqrt{}}{2} $
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