(3x-4)(6-x)=0

Solution for (3x-4)(6-x)=0 equation:

(3x-4)(6-x)=0

We add all the numbers together, and all the variables

(3x-4)(-1x+6)=0

We multiply parentheses ..

(-3x^2+18x+4x-24)=0

We get rid of parentheses

-3x^2+18x+4x-24=0

We add all the numbers together, and all the variables

-3x^2+22x-24=0

a = -3; b = 22; c = -24;
Δ = b2-4ac
Δ = 222-4·(-3)·(-24)
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-14}{2*-3}=\frac{-36}{-6}
=+6
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+14}{2*-3}=\frac{-8}{-6}
=1+1/3
$