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4f^2=64
We move all terms to the left:
4f^2-(64)=0
a = 4; b = 0; c = -64;
Δ = b2-4ac
Δ = 02-4·4·(-64)
Δ = 1024
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{1024}=32$$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-32}{2*4}=\frac{-32}{8} =-4 $$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+32}{2*4}=\frac{32}{8} =4 $
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