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(F)=4F+4/3F-11
We move all terms to the left:
(F)-(4F+4/3F-11)=0
Domain of the equation: 3F-11)!=0We get rid of parentheses
F∈R
F-4F-4/3F+11=0
We multiply all the terms by the denominator
F*3F-4F*3F+11*3F-4=0
Wy multiply elements
3F^2-12F^2+33F-4=0
We add all the numbers together, and all the variables
-9F^2+33F-4=0
a = -9; b = 33; c = -4;
Δ = b2-4ac
Δ = 332-4·(-9)·(-4)
Δ = 945
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}$$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{945}=\sqrt{9*105}=\sqrt{9}*\sqrt{105}=3\sqrt{105}$$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(33)-3\sqrt{105}}{2*-9}=\frac{-33-3\sqrt{105}}{-18} $$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(33)+3\sqrt{105}}{2*-9}=\frac{-33+3\sqrt{105}}{-18} $
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