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(F)=5F^2+3F-1
We move all terms to the left:
(F)-(5F^2+3F-1)=0
We get rid of parentheses
-5F^2+F-3F+1=0
We add all the numbers together, and all the variables
-5F^2-2F+1=0
a = -5; b = -2; c = +1;
Δ = b2-4ac
Δ = -22-4·(-5)·1
Δ = 24
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}$$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{24}=\sqrt{4*6}=\sqrt{4}*\sqrt{6}=2\sqrt{6}$$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{6}}{2*-5}=\frac{2-2\sqrt{6}}{-10} $$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{6}}{2*-5}=\frac{2+2\sqrt{6}}{-10} $
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