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f2+64=353
We move all terms to the left:
f2+64-(353)=0
We add all the numbers together, and all the variables
f^2-289=0
a = 1; b = 0; c = -289;
Δ = b2-4ac
Δ = 02-4·1·(-289)
Δ = 1156
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}$$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{1156}=34$$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-34}{2*1}=\frac{-34}{2} =-17 $$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+34}{2*1}=\frac{34}{2} =17 $
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