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(0)=6F^2-9F+3
We move all terms to the left:
(0)-(6F^2-9F+3)=0
We add all the numbers together, and all the variables
-(6F^2-9F+3)=0
We get rid of parentheses
-6F^2+9F-3=0
a = -6; b = 9; c = -3;
Δ = b2-4ac
Δ = 92-4·(-6)·(-3)
Δ = 9
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{9}=3$$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-3}{2*-6}=\frac{-12}{-12} =1 $$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+3}{2*-6}=\frac{-6}{-12} =1/2 $
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