20=1+2(x^2)

Solution for 20=1+2(x^2) equation:

20=1+2(x^2)

We move all terms to the left:

20-(1+2(x^2))=0

We get rid of parentheses

-2x^2-1+20=0

We add all the numbers together, and all the variables

-2x^2+19=0

a = -2; b = 0; c = +19;
Δ = b2-4ac
Δ = 02-4·(-2)·19
Δ = 152
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{152}=\sqrt{4*38}=\sqrt{4}*\sqrt{38}=2\sqrt{38}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{38}}{2*-2}=\frac{0-2\sqrt{38}}{-4}
=-\frac{2\sqrt{38}}{-4}
=-\frac{\sqrt{38}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{38}}{2*-2}=\frac{0+2\sqrt{38}}{-4}
=\frac{2\sqrt{38}}{-4}
=\frac{\sqrt{38}}{-2}
$