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(F)=2/5F+8
We move all terms to the left:
(F)-(2/5F+8)=0
Domain of the equation: 5F+8)!=0We get rid of parentheses
F∈R
F-2/5F-8=0
We multiply all the terms by the denominator
F*5F-8*5F-2=0
Wy multiply elements
5F^2-40F-2=0
a = 5; b = -40; c = -2;
Δ = b2-4ac
Δ = -402-4·5·(-2)
Δ = 1640
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}$$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1640}=\sqrt{4*410}=\sqrt{4}*\sqrt{410}=2\sqrt{410}$$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-2\sqrt{410}}{2*5}=\frac{40-2\sqrt{410}}{10} $$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+2\sqrt{410}}{2*5}=\frac{40+2\sqrt{410}}{10} $
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