1.04=(1+x)+x(1+x)

Solution for 1.04=(1+x)+x(1+x) equation:

1.04=(1+x)+x(1+x)

We move all terms to the left:

1.04-((1+x)+x(1+x))=0

We add all the numbers together, and all the variables

-((x+1)+x(x+1))+1.04=0


We calculate terms in parentheses: -((x+1)+x(x+1)), so:

(x+1)+x(x+1)

We multiply parentheses

x^2+(x+1)+x

We get rid of parentheses

x^2+x+x+1

We add all the numbers together, and all the variables

x^2+2x+1

Back to the equation:

-(x^2+2x+1)


We get rid of parentheses

-x^2-2x-1+1.04=0

We add all the numbers together, and all the variables

-1x^2-2x+0.04=0

a = -1; b = -2; c = +0.04;
Δ = b2-4ac
Δ = -22-4·(-1)·0.04
Δ = 4.16
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{-(-2)-\sqrt{4.16}}{2*-1}=\frac{2-\sqrt{4.16}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+\sqrt{4.16}}{2*-1}=\frac{2+\sqrt{4.16}}{-2}
$