x=(1+0.05x)(40+20x)

Solution for x=(1+0.05x)(40+20x) equation:

x=(1+0.05x)(40+20x)

We move all terms to the left:

x-((1+0.05x)(40+20x))=0

We add all the numbers together, and all the variables

x-((0.05x+1)(20x+40))=0

We multiply parentheses ..

-((+0x^2+0x+20x+40))+x=0


We calculate terms in parentheses: -((+0x^2+0x+20x+40)), so:

(+0x^2+0x+20x+40)

We get rid of parentheses

0x^2+0x+20x+40

We add all the numbers together, and all the variables

x^2+21x+40

Back to the equation:

-(x^2+21x+40)


We add all the numbers together, and all the variables

x-(x^2+21x+40)=0

We get rid of parentheses

-x^2+x-21x-40=0

We add all the numbers together, and all the variables

-1x^2-20x-40=0

a = -1; b = -20; c = -40;
Δ = b2-4ac
Δ = -202-4·(-1)·(-40)
Δ = 240
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:

$\sqrt{\Delta}=\sqrt{240}=\sqrt{16*15}=\sqrt{16}*\sqrt{15}=4\sqrt{15}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-4\sqrt{15}}{2*-1}=\frac{20-4\sqrt{15}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4\sqrt{15}}{2*-1}=\frac{20+4\sqrt{15}}{-2}
$