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(F)=9F^2+3F
We move all terms to the left:
(F)-(9F^2+3F)=0
We get rid of parentheses
-9F^2+F-3F=0
We add all the numbers together, and all the variables
-9F^2-2F=0
a = -9; b = -2; c = 0;
Δ = b2-4ac
Δ = -22-4·(-9)·0
Δ = 4
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{4}=2$$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2}{2*-9}=\frac{0}{-18} =0 $$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2}{2*-9}=\frac{4}{-18} =-2/9 $
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