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4f^2=324
We move all terms to the left:
4f^2-(324)=0
a = 4; b = 0; c = -324;
Δ = b2-4ac
Δ = 02-4·4·(-324)
Δ = 5184
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{5184}=72$$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-72}{2*4}=\frac{-72}{8} =-9 $$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+72}{2*4}=\frac{72}{8} =9 $
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