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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(-1x^2+40x+5x-200)=0

We get rid of parentheses

-1x^2+40x+5x-200=0

We add all the numbers together, and all the variables

-1x^2+45x-200=0

a = -1; b = 45; c = -200;
Δ = b2-4ac
Δ = 452-4·(-1)·(-200)
Δ = 1225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1225}=35$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(45)-35}{2*-1}=\frac{-80}{-2}
=+40
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(45)+35}{2*-1}=\frac{-10}{-2}
=+5
$