F(1)=3,f(2)=1

Solution for F(1)=3,f(2)=1 equation:

(1)=3.F(2)=1

We move all terms to the left:

(1)-(3.F(2))=0

We add all the numbers together, and all the variables

-(+3.F^2)+1=0

We get rid of parentheses

-3.F^2+1=0

We add all the numbers together, and all the variables

-3F^2+1=0

a = -3; b = 0; c = +1;
Δ = b2-4ac
Δ = 02-4·(-3)·1
Δ = 12
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:

$\sqrt{\Delta}=\sqrt{12}=\sqrt{4*3}=\sqrt{4}*\sqrt{3}=2\sqrt{3}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{3}}{2*-3}=\frac{0-2\sqrt{3}}{-6}
=-\frac{2\sqrt{3}}{-6}
=-\frac{\sqrt{3}}{-3}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{3}}{2*-3}=\frac{0+2\sqrt{3}}{-6}
=\frac{2\sqrt{3}}{-6}
=\frac{\sqrt{3}}{-3}
$