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(F)=-3F^2+5F+1
We move all terms to the left:
(F)-(-3F^2+5F+1)=0
We get rid of parentheses
3F^2-5F+F-1=0
We add all the numbers together, and all the variables
3F^2-4F-1=0
a = 3; b = -4; c = -1;
Δ = b2-4ac
Δ = -42-4·3·(-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{-(-4)-2\sqrt{7}}{2*3}=\frac{4-2\sqrt{7}}{6} $$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2\sqrt{7}}{2*3}=\frac{4+2\sqrt{7}}{6} $
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