g0+g1+g2=(1+g)2

Solution for g0+g1+g2=(1+g)2 equation:

g0+g1+g2=(1+g)2

We move all terms to the left:

g0+g1+g2-((1+g)2)=0

We add all the numbers together, and all the variables

g0+g1+g2-((g+1)2)=0

We add all the numbers together, and all the variables

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


We calculate terms in parentheses: -((g+1)2), so:

(g+1)2

We multiply parentheses

2g+2

Back to the equation:

-(2g+2)


We get rid of parentheses

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

We add all the numbers together, and all the variables

g^2-2=0

a = 1; b = 0; c = -2;
Δ = b2-4ac
Δ = 02-4·1·(-2)
Δ = 8
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{8}=\sqrt{4*2}=\sqrt{4}*\sqrt{2}=2\sqrt{2}$
$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{2}}{2*1}=\frac{0-2\sqrt{2}}{2}
=-\frac{2\sqrt{2}}{2}
=-\sqrt{2}
$
$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{2}}{2*1}=\frac{0+2\sqrt{2}}{2}
=\frac{2\sqrt{2}}{2}
=\sqrt{2}
$