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(F)=-2/3F+5
We move all terms to the left:
(F)-(-2/3F+5)=0
Domain of the equation: 3F+5)!=0We get rid of parentheses
F∈R
F+2/3F-5=0
We multiply all the terms by the denominator
F*3F-5*3F+2=0
Wy multiply elements
3F^2-15F+2=0
a = 3; b = -15; c = +2;
Δ = b2-4ac
Δ = -152-4·3·2
Δ = 201
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}$$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-\sqrt{201}}{2*3}=\frac{15-\sqrt{201}}{6} $$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+\sqrt{201}}{2*3}=\frac{15+\sqrt{201}}{6} $
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