40=40x(20+x)

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

40=40x(20+x)

We move all terms to the left:

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

We add all the numbers together, and all the variables

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


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

40x(x+20)

We multiply parentheses

40x^2+800x

Back to the equation:

-(40x^2+800x)


We get rid of parentheses

-40x^2-800x+40=0

a = -40; b = -800; c = +40;
Δ = b2-4ac
Δ = -8002-4·(-40)·40
Δ = 646400
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:

$\sqrt{\Delta}=\sqrt{646400}=\sqrt{6400*101}=\sqrt{6400}*\sqrt{101}=80\sqrt{101}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-800)-80\sqrt{101}}{2*-40}=\frac{800-80\sqrt{101}}{-80}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-800)+80\sqrt{101}}{2*-40}=\frac{800+80\sqrt{101}}{-80}
$