(3x-4)(x-8)=0

Solution for (3x-4)(x-8)=0 equation:

(3x-4)(x-8)=0

We multiply parentheses ..

(+3x^2-24x-4x+32)=0

We get rid of parentheses

3x^2-24x-4x+32=0

We add all the numbers together, and all the variables

3x^2-28x+32=0

a = 3; b = -28; c = +32;
Δ = b2-4ac
Δ = -282-4·3·32
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-20}{2*3}=\frac{8}{6}
=1+1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+20}{2*3}=\frac{48}{6}
=8
$