20x^2+1=3

Solution for 20x^2+1=3 equation:

20x^2+1=3

We move all terms to the left:

20x^2+1-(3)=0

We add all the numbers together, and all the variables

20x^2-2=0

a = 20; b = 0; c = -2;
Δ = b2-4ac
Δ = 02-4·20·(-2)
Δ = 160
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:

$\sqrt{\Delta}=\sqrt{160}=\sqrt{16*10}=\sqrt{16}*\sqrt{10}=4\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{10}}{2*20}=\frac{0-4\sqrt{10}}{40}
=-\frac{4\sqrt{10}}{40}
=-\frac{\sqrt{10}}{10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{10}}{2*20}=\frac{0+4\sqrt{10}}{40}
=\frac{4\sqrt{10}}{40}
=\frac{\sqrt{10}}{10}
$