(4x-5)(10-x)=0

Solution for (4x-5)(10-x)=0 equation:

(4x-5)(10-x)=0

We add all the numbers together, and all the variables

(4x-5)(-1x+10)=0

We multiply parentheses ..

(-4x^2+40x+5x-50)=0

We get rid of parentheses

-4x^2+40x+5x-50=0

We add all the numbers together, and all the variables

-4x^2+45x-50=0

a = -4; b = 45; c = -50;
Δ = b2-4ac
Δ = 452-4·(-4)·(-50)
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*-4}=\frac{-80}{-8}
=+10
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(45)+35}{2*-4}=\frac{-10}{-8}
=1+1/4
$