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