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f(2)=9-2^2
We move all terms to the left:
f(2)-(9-2^2)=0
We add all the numbers together, and all the variables
f^2-5=0
a = 1; b = 0; c = -5;
Δ = b2-4ac
Δ = 02-4·1·(-5)
Δ = 20
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}$$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{20}=\sqrt{4*5}=\sqrt{4}*\sqrt{5}=2\sqrt{5}$$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{5}}{2*1}=\frac{0-2\sqrt{5}}{2} =-\frac{2\sqrt{5}}{2} =-\sqrt{5} $$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{5}}{2*1}=\frac{0+2\sqrt{5}}{2} =\frac{2\sqrt{5}}{2} =\sqrt{5} $
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