F(1)=6,f(2)=-6,f(3)=1

Solution for F(1)=6,f(2)=-6,f(3)=1 equation:

(1)=6.F(2)=-6.F(3)=1

We move all terms to the left:

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

We add all the numbers together, and all the variables

-(+6.F^2)+1=0

We get rid of parentheses

-6.F^2+1=0

We add all the numbers together, and all the variables

-6F^2+1=0

a = -6; b = 0; c = +1;
Δ = b2-4ac
Δ = 02-4·(-6)·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{-(0)-2\sqrt{6}}{2*-6}=\frac{0-2\sqrt{6}}{-12}
=-\frac{2\sqrt{6}}{-12}
=-\frac{\sqrt{6}}{-6}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{6}}{2*-6}=\frac{0+2\sqrt{6}}{-12}
=\frac{2\sqrt{6}}{-12}
=\frac{\sqrt{6}}{-6}
$