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(F)=5/2F-3
We move all terms to the left:
(F)-(5/2F-3)=0
Domain of the equation: 2F-3)!=0We get rid of parentheses
F∈R
F-5/2F+3=0
We multiply all the terms by the denominator
F*2F+3*2F-5=0
Wy multiply elements
2F^2+6F-5=0
a = 2; b = 6; c = -5;
Δ = b2-4ac
Δ = 62-4·2·(-5)
Δ = 76
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}$$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{76}=\sqrt{4*19}=\sqrt{4}*\sqrt{19}=2\sqrt{19}$$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{19}}{2*2}=\frac{-6-2\sqrt{19}}{4} $$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{19}}{2*2}=\frac{-6+2\sqrt{19}}{4} $
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