(4-3x)(6-5x)=0

Solution for (4-3x)(6-5x)=0 equation:

(4-3x)(6-5x)=0

We add all the numbers together, and all the variables

(-3x+4)(-5x+6)=0

We multiply parentheses ..

(+15x^2-18x-20x+24)=0

We get rid of parentheses

15x^2-18x-20x+24=0

We add all the numbers together, and all the variables

15x^2-38x+24=0

a = 15; b = -38; c = +24;
Δ = b2-4ac
Δ = -382-4·15·24
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-38)-2}{2*15}=\frac{36}{30}
=1+1/5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-38)+2}{2*15}=\frac{40}{30}
=1+1/3
$