100x^2+20x+1=16

Solution for 100x^2+20x+1=16 equation:

100x^2+20x+1=16

We move all terms to the left:

100x^2+20x+1-(16)=0

We add all the numbers together, and all the variables

100x^2+20x-15=0

a = 100; b = 20; c = -15;
Δ = b2-4ac
Δ = 202-4·100·(-15)
Δ = 6400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{6400}=80$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-80}{2*100}=\frac{-100}{200}
=-1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+80}{2*100}=\frac{60}{200}
=3/10
$