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(F)=(3F+1)(2F-1)
We move all terms to the left:
(F)-((3F+1)(2F-1))=0
We multiply parentheses ..
-((+6F^2-3F+2F-1))+F=0
We calculate terms in parentheses: -((+6F^2-3F+2F-1)), so:We add all the numbers together, and all the variables
(+6F^2-3F+2F-1)
We get rid of parentheses
6F^2-3F+2F-1
We add all the numbers together, and all the variables
6F^2-1F-1
Back to the equation:
-(6F^2-1F-1)
F-(6F^2-1F-1)=0
We get rid of parentheses
-6F^2+F+1F+1=0
We add all the numbers together, and all the variables
-6F^2+2F+1=0
a = -6; b = 2; c = +1;
Δ = b2-4ac
Δ = 22-4·(-6)·1
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{-(2)-2\sqrt{7}}{2*-6}=\frac{-2-2\sqrt{7}}{-12} $$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{7}}{2*-6}=\frac{-2+2\sqrt{7}}{-12} $
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