F(1)=4x^2-2x

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Solution for F(1)=4x^2-2x equation:



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

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