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

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

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

We multiply parentheses ..

(+12x^2-16x-15x+20)=0

We get rid of parentheses

12x^2-16x-15x+20=0

We add all the numbers together, and all the variables

12x^2-31x+20=0

a = 12; b = -31; c = +20;
Δ = b2-4ac
Δ = -312-4·12·20
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-31)-1}{2*12}=\frac{30}{24}
=1+1/4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-31)+1}{2*12}=\frac{32}{24}
=1+1/3
$