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f(2)=3^2-2*2+6
We move all terms to the left:
f(2)-(3^2-2*2+6)=0
We add all the numbers together, and all the variables
f^2-11=0
a = 1; b = 0; c = -11;
Δ = b2-4ac
Δ = 02-4·1·(-11)
Δ = 44
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}$$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{44}=\sqrt{4*11}=\sqrt{4}*\sqrt{11}=2\sqrt{11}$$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{11}}{2*1}=\frac{0-2\sqrt{11}}{2} =-\frac{2\sqrt{11}}{2} =-\sqrt{11} $$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{11}}{2*1}=\frac{0+2\sqrt{11}}{2} =\frac{2\sqrt{11}}{2} =\sqrt{11} $
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