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(F)=6F^2-9F+3
We move all terms to the left:
(F)-(6F^2-9F+3)=0
We get rid of parentheses
-6F^2+F+9F-3=0
We add all the numbers together, and all the variables
-6F^2+10F-3=0
a = -6; b = 10; c = -3;
Δ = b2-4ac
Δ = 102-4·(-6)·(-3)
Δ = 28
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}$$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{28}=\sqrt{4*7}=\sqrt{4}*\sqrt{7}=2\sqrt{7}$$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{7}}{2*-6}=\frac{-10-2\sqrt{7}}{-12} $$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{7}}{2*-6}=\frac{-10+2\sqrt{7}}{-12} $
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