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(F)=F2+3F-4
We move all terms to the left:
(F)-(F2+3F-4)=0
We add all the numbers together, and all the variables
-(+F^2+3F-4)+F=0
We get rid of parentheses
-F^2-3F+F+4=0
We add all the numbers together, and all the variables
-1F^2-2F+4=0
a = -1; b = -2; c = +4;
Δ = b2-4ac
Δ = -22-4·(-1)·4
Δ = 20
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}$$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{20}=\sqrt{4*5}=\sqrt{4}*\sqrt{5}=2\sqrt{5}$$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{5}}{2*-1}=\frac{2-2\sqrt{5}}{-2} $$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{5}}{2*-1}=\frac{2+2\sqrt{5}}{-2} $
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