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(F)=F2-5F-3
We move all terms to the left:
(F)-(F2-5F-3)=0
We add all the numbers together, and all the variables
-(+F^2-5F-3)+F=0
We get rid of parentheses
-F^2+5F+F+3=0
We add all the numbers together, and all the variables
-1F^2+6F+3=0
a = -1; b = 6; c = +3;
Δ = b2-4ac
Δ = 62-4·(-1)·3
Δ = 48
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}$$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-4\sqrt{3}}{2*-1}=\frac{-6-4\sqrt{3}}{-2} $$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+4\sqrt{3}}{2*-1}=\frac{-6+4\sqrt{3}}{-2} $
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