(7x-3)(4x-1)=0

Solution for (7x-3)(4x-1)=0 equation:

(7x-3)(4x-1)=0

We multiply parentheses ..

(+28x^2-7x-12x+3)=0

We get rid of parentheses

28x^2-7x-12x+3=0

We add all the numbers together, and all the variables

28x^2-19x+3=0

a = 28; b = -19; c = +3;
Δ = b2-4ac
Δ = -192-4·28·3
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-5}{2*28}=\frac{14}{56}
=1/4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+5}{2*28}=\frac{24}{56}
=3/7
$