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f(2)+1=3((2)+1)+2
We move all terms to the left:
f(2)+1-(3((2)+1)+2)=0
We add all the numbers together, and all the variables
f2+1-(33+2)=0
We add all the numbers together, and all the variables
f^2-34=0
a = 1; b = 0; c = -34;
Δ = b2-4ac
Δ = 02-4·1·(-34)
Δ = 136
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}$$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{136}=\sqrt{4*34}=\sqrt{4}*\sqrt{34}=2\sqrt{34}$$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{34}}{2*1}=\frac{0-2\sqrt{34}}{2} =-\frac{2\sqrt{34}}{2} =-\sqrt{34} $$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{34}}{2*1}=\frac{0+2\sqrt{34}}{2} =\frac{2\sqrt{34}}{2} =\sqrt{34} $
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