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(3)=-4F^2-2F+5
We move all terms to the left:
(3)-(-4F^2-2F+5)=0
We get rid of parentheses
4F^2+2F-5+3=0
We add all the numbers together, and all the variables
4F^2+2F-2=0
a = 4; b = 2; c = -2;
Δ = b2-4ac
Δ = 22-4·4·(-2)
Δ = 36
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}$$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{36}=6$$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-6}{2*4}=\frac{-8}{8} =-1 $$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+6}{2*4}=\frac{4}{8} =1/2 $
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