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(F)=5/5F-2
We move all terms to the left:
(F)-(5/5F-2)=0
Domain of the equation: 5F-2)!=0We get rid of parentheses
F∈R
F-5/5F+2=0
We multiply all the terms by the denominator
F*5F+2*5F-5=0
Wy multiply elements
5F^2+10F-5=0
a = 5; b = 10; c = -5;
Δ = b2-4ac
Δ = 102-4·5·(-5)
Δ = 200
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}$$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{200}=\sqrt{100*2}=\sqrt{100}*\sqrt{2}=10\sqrt{2}$$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-10\sqrt{2}}{2*5}=\frac{-10-10\sqrt{2}}{10} $$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+10\sqrt{2}}{2*5}=\frac{-10+10\sqrt{2}}{10} $
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