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(F)=-1/4F+3
We move all terms to the left:
(F)-(-1/4F+3)=0
Domain of the equation: 4F+3)!=0We get rid of parentheses
F∈R
F+1/4F-3=0
We multiply all the terms by the denominator
F*4F-3*4F+1=0
Wy multiply elements
4F^2-12F+1=0
a = 4; b = -12; c = +1;
Δ = b2-4ac
Δ = -122-4·4·1
Δ = 128
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}$$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{128}=\sqrt{64*2}=\sqrt{64}*\sqrt{2}=8\sqrt{2}$$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-8\sqrt{2}}{2*4}=\frac{12-8\sqrt{2}}{8} $$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+8\sqrt{2}}{2*4}=\frac{12+8\sqrt{2}}{8} $
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