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40=f(2)
We move all terms to the left:
40-(f(2))=0
determiningTheFunctionDomain -f2+40=0
We add all the numbers together, and all the variables
-1f^2+40=0
a = -1; b = 0; c = +40;
Δ = b2-4ac
Δ = 02-4·(-1)·40
Δ = 160
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}$$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{160}=\sqrt{16*10}=\sqrt{16}*\sqrt{10}=4\sqrt{10}$$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{10}}{2*-1}=\frac{0-4\sqrt{10}}{-2} =-\frac{4\sqrt{10}}{-2} =-\frac{2\sqrt{10}}{-1} $$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{10}}{2*-1}=\frac{0+4\sqrt{10}}{-2} =\frac{4\sqrt{10}}{-2} =\frac{2\sqrt{10}}{-1} $
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