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f(3-1)=24
We move all terms to the left:
f(3-1)-(24)=0
We add all the numbers together, and all the variables
f2-24=0
We add all the numbers together, and all the variables
f^2-24=0
a = 1; b = 0; c = -24;
Δ = b2-4ac
Δ = 02-4·1·(-24)
Δ = 96
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}$$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{96}=\sqrt{16*6}=\sqrt{16}*\sqrt{6}=4\sqrt{6}$$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{6}}{2*1}=\frac{0-4\sqrt{6}}{2} =-\frac{4\sqrt{6}}{2} =-2\sqrt{6} $$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{6}}{2*1}=\frac{0+4\sqrt{6}}{2} =\frac{4\sqrt{6}}{2} =2\sqrt{6} $
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