F(x+3)=x^2-4x+6

Solution for F(x+3)=x^2-4x+6 equation:

(F+3)=F^2-4F+6

We move all terms to the left:

(F+3)-(F^2-4F+6)=0

We get rid of parentheses

-F^2+F+4F+3-6=0

We add all the numbers together, and all the variables

-1F^2+5F-3=0

a = -1; b = 5; c = -3;
Δ = b2-4ac
Δ = 52-4·(-1)·(-3)
Δ = 13
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-\sqrt{13}}{2*-1}=\frac{-5-\sqrt{13}}{-2}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{13}}{2*-1}=\frac{-5+\sqrt{13}}{-2}
$