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(F)=3F^2-5F+2F=2
We move all terms to the left:
(F)-(3F^2-5F+2F)=0
We get rid of parentheses
-3F^2+F+5F-2F=0
We add all the numbers together, and all the variables
-3F^2+4F=0
a = -3; b = 4; c = 0;
Δ = b2-4ac
Δ = 42-4·(-3)·0
Δ = 16
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}$$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{16}=4$$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4}{2*-3}=\frac{-8}{-6} =1+1/3 $$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4}{2*-3}=\frac{0}{-6} =0 $
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