5*10^(2x)+1=21

Solution for 5*10^(2x)+1=21 equation:

5*10^(2x)+1=21

We move all terms to the left:

5*10^(2x)+1-(21)=0

We add all the numbers together, and all the variables

5*10^2x-20=0

Wy multiply elements

50x^2-20=0

a = 50; b = 0; c = -20;
Δ = b2-4ac
Δ = 02-4·50·(-20)
Δ = 4000
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:

$\sqrt{\Delta}=\sqrt{4000}=\sqrt{400*10}=\sqrt{400}*\sqrt{10}=20\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-20\sqrt{10}}{2*50}=\frac{0-20\sqrt{10}}{100}
=-\frac{20\sqrt{10}}{100}
=-\frac{\sqrt{10}}{5}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+20\sqrt{10}}{2*50}=\frac{0+20\sqrt{10}}{100}
=\frac{20\sqrt{10}}{100}
=\frac{\sqrt{10}}{5}
$