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16f^2-80=0
a = 16; b = 0; c = -80;
Δ = b2-4ac
Δ = 02-4·16·(-80)
Δ = 5120
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}$$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{5120}=\sqrt{1024*5}=\sqrt{1024}*\sqrt{5}=32\sqrt{5}$$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-32\sqrt{5}}{2*16}=\frac{0-32\sqrt{5}}{32} =-\frac{32\sqrt{5}}{32} =-\sqrt{5} $$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+32\sqrt{5}}{2*16}=\frac{0+32\sqrt{5}}{32} =\frac{32\sqrt{5}}{32} =\sqrt{5} $
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