X^2+20x=x+1

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

X^2+20X=X+1

We move all terms to the left:

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

We get rid of parentheses

X^2+20X-X-1=0

We add all the numbers together, and all the variables

X^2+19X-1=0

a = 1; b = 19; c = -1;
Δ = b2-4ac
Δ = 192-4·1·(-1)
Δ = 365
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-\sqrt{365}}{2*1}=\frac{-19-\sqrt{365}}{2}
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+\sqrt{365}}{2*1}=\frac{-19+\sqrt{365}}{2}
$