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(F)=-3F^2-5F+13
We move all terms to the left:
(F)-(-3F^2-5F+13)=0
We get rid of parentheses
3F^2+5F+F-13=0
We add all the numbers together, and all the variables
3F^2+6F-13=0
a = 3; b = 6; c = -13;
Δ = b2-4ac
Δ = 62-4·3·(-13)
Δ = 192
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}$$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{192}=\sqrt{64*3}=\sqrt{64}*\sqrt{3}=8\sqrt{3}$$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-8\sqrt{3}}{2*3}=\frac{-6-8\sqrt{3}}{6} $$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+8\sqrt{3}}{2*3}=\frac{-6+8\sqrt{3}}{6} $
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