g0+g1+g2=(1+g)

Solution for g0+g1+g2=(1+g) equation:

g0+g1+g2=(1+g)

We move all terms to the left:

g0+g1+g2-((1+g))=0

We add all the numbers together, and all the variables

g0+g1+g2-((g+1))=0

We add all the numbers together, and all the variables

g^2+2g-((g+1))=0


We calculate terms in parentheses: -((g+1)), so:

(g+1)

We get rid of parentheses

g+1

Back to the equation:

-(g+1)


We get rid of parentheses

g^2+2g-g-1=0

We add all the numbers together, and all the variables

g^2+g-1=0

a = 1; b = 1; c = -1;
Δ = b2-4ac
Δ = 12-4·1·(-1)
Δ = 5
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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