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f^2=154
We move all terms to the left:
f^2-(154)=0
a = 1; b = 0; c = -154;
Δ = b2-4ac
Δ = 02-4·1·(-154)
Δ = 616
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}$$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{616}=\sqrt{4*154}=\sqrt{4}*\sqrt{154}=2\sqrt{154}$$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{154}}{2*1}=\frac{0-2\sqrt{154}}{2} =-\frac{2\sqrt{154}}{2} =-\sqrt{154} $$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{154}}{2*1}=\frac{0+2\sqrt{154}}{2} =\frac{2\sqrt{154}}{2} =\sqrt{154} $
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