F(1+2)=2f(1+1)+2f(1)

Solution for F(1+2)=2f(1+1)+2f(1) equation:

(1+2)=2F(1+1)+2F(1)

We move all terms to the left:

(1+2)-(2F(1+1)+2F(1))=0

We add all the numbers together, and all the variables

-(2F2+2F1)+3=0

We get rid of parentheses

-2F2-2F1+3=0

We add all the numbers together, and all the variables

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

a = -2; b = -2; c = +3;
Δ = b2-4ac
Δ = -22-4·(-2)·3
Δ = 28
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:

$\sqrt{\Delta}=\sqrt{28}=\sqrt{4*7}=\sqrt{4}*\sqrt{7}=2\sqrt{7}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{7}}{2*-2}=\frac{2-2\sqrt{7}}{-4}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{7}}{2*-2}=\frac{2+2\sqrt{7}}{-4}
$