(4x-3)(5x=1)=(2x-4)(10x+3)

Solution for (4x-3)(5x=1)=(2x-4)(10x+3) equation:

(4x-3)(5x=1)=(2x-4)(10x+3)

We move all terms to the left:

(4x-3)(5x-(1))=0

We multiply parentheses ..

(+20x^2-4x-15x+3)=0

We get rid of parentheses

20x^2-4x-15x+3=0

We add all the numbers together, and all the variables

20x^2-19x+3=0

a = 20; b = -19; c = +3;
Δ = b2-4ac
Δ = -192-4·20·3
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{-(-19)-11}{2*20}=\frac{8}{40}
=1/5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+11}{2*20}=\frac{30}{40}
=3/4
$