G(2c)=(2c)^2-(2C)

Solution for G(2c)=(2c)^2-(2C) equation:

(2G)=(2G)^2-(2)

We move all terms to the left:

(2G)-((2G)^2-(2))=0

We get rid of parentheses

-2G^2+2G+2=0

a = -2; b = 2; c = +2;
Δ = b2-4ac
Δ = 22-4·(-2)·2
Δ = 20
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:

$\sqrt{\Delta}=\sqrt{20}=\sqrt{4*5}=\sqrt{4}*\sqrt{5}=2\sqrt{5}$
$G_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{5}}{2*-2}=\frac{-2-2\sqrt{5}}{-4}
$
$G_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{5}}{2*-2}=\frac{-2+2\sqrt{5}}{-4}
$