(3x-5)(4-x)=0

Solution for (3x-5)(4-x)=0 equation:

(3x-5)(4-x)=0

We add all the numbers together, and all the variables

(3x-5)(-1x+4)=0

We multiply parentheses ..

(-3x^2+12x+5x-20)=0

We get rid of parentheses

-3x^2+12x+5x-20=0

We add all the numbers together, and all the variables

-3x^2+17x-20=0

a = -3; b = 17; c = -20;
Δ = b2-4ac
Δ = 172-4·(-3)·(-20)
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-7}{2*-3}=\frac{-24}{-6}
=+4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+7}{2*-3}=\frac{-10}{-6}
=1+2/3
$