(40+1x)(200-5x)=0

Solution for (40+1x)(200-5x)=0 equation:

(40+1x)(200-5x)=0

We add all the numbers together, and all the variables

(x+40)(-5x+200)=0

We multiply parentheses ..

(-5x^2+200x-200x+8000)=0

We get rid of parentheses

-5x^2+200x-200x+8000=0

We add all the numbers together, and all the variables

-5x^2+8000=0

a = -5; b = 0; c = +8000;
Δ = b2-4ac
Δ = 02-4·(-5)·8000
Δ = 160000
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{160000}=400$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-400}{2*-5}=\frac{-400}{-10}
=+40
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+400}{2*-5}=\frac{400}{-10}
=-40
$