G(17)=2x^2-4

Solution for G(17)=2x^2-4 equation:

(17)=2G^2-4

We move all terms to the left:

(17)-(2G^2-4)=0

We get rid of parentheses

-2G^2+4+17=0

We add all the numbers together, and all the variables

-2G^2+21=0

a = -2; b = 0; c = +21;
Δ = b2-4ac
Δ = 02-4·(-2)·21
Δ = 168
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:

$\sqrt{\Delta}=\sqrt{168}=\sqrt{4*42}=\sqrt{4}*\sqrt{42}=2\sqrt{42}$
$G_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{42}}{2*-2}=\frac{0-2\sqrt{42}}{-4}
=-\frac{2\sqrt{42}}{-4}
=-\frac{\sqrt{42}}{-2}
$
$G_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{42}}{2*-2}=\frac{0+2\sqrt{42}}{-4}
=\frac{2\sqrt{42}}{-4}
=\frac{\sqrt{42}}{-2}
$