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

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

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

We move all terms to the left:

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

We multiply parentheses ..

(+15x^2-9x-20x+12)=0

We get rid of parentheses

15x^2-9x-20x+12=0

We add all the numbers together, and all the variables

15x^2-29x+12=0

a = 15; b = -29; c = +12;
Δ = b2-4ac
Δ = -292-4·15·12
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-29)-11}{2*15}=\frac{18}{30}
=3/5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-29)+11}{2*15}=\frac{40}{30}
=1+1/3
$