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(F)=(F^2-3F-6)
We move all terms to the left:
(F)-((F^2-3F-6))=0
We calculate terms in parentheses: -((F^2-3F-6)), so:We get rid of parentheses
(F^2-3F-6)
We get rid of parentheses
F^2-3F-6
Back to the equation:
-(F^2-3F-6)
-F^2+F+3F+6=0
We add all the numbers together, and all the variables
-1F^2+4F+6=0
a = -1; b = 4; c = +6;
Δ = b2-4ac
Δ = 42-4·(-1)·6
Δ = 40
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}$$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{40}=\sqrt{4*10}=\sqrt{4}*\sqrt{10}=2\sqrt{10}$$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{10}}{2*-1}=\frac{-4-2\sqrt{10}}{-2} $$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{10}}{2*-1}=\frac{-4+2\sqrt{10}}{-2} $
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