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(F)=9F^2+2F-3
We move all terms to the left:
(F)-(9F^2+2F-3)=0
We get rid of parentheses
-9F^2+F-2F+3=0
We add all the numbers together, and all the variables
-9F^2-1F+3=0
a = -9; b = -1; c = +3;
Δ = b2-4ac
Δ = -12-4·(-9)·3
Δ = 109
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{-(-1)-\sqrt{109}}{2*-9}=\frac{1-\sqrt{109}}{-18} $$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{109}}{2*-9}=\frac{1+\sqrt{109}}{-18} $
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