20*2^2x+1=500

Solution for 20*2^2x+1=500 equation:

20*2^2x+1=500

We move all terms to the left:

20*2^2x+1-(500)=0

We add all the numbers together, and all the variables

20*2^2x-499=0

Wy multiply elements

40x^2-499=0

a = 40; b = 0; c = -499;
Δ = b2-4ac
Δ = 02-4·40·(-499)
Δ = 79840
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:

$\sqrt{\Delta}=\sqrt{79840}=\sqrt{16*4990}=\sqrt{16}*\sqrt{4990}=4\sqrt{4990}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{4990}}{2*40}=\frac{0-4\sqrt{4990}}{80}
=-\frac{4\sqrt{4990}}{80}
=-\frac{\sqrt{4990}}{20}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{4990}}{2*40}=\frac{0+4\sqrt{4990}}{80}
=\frac{4\sqrt{4990}}{80}
=\frac{\sqrt{4990}}{20}
$