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

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

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

We multiply parentheses ..

(+28x^2+4x-21x-3)=0

We get rid of parentheses

28x^2+4x-21x-3=0

We add all the numbers together, and all the variables

28x^2-17x-3=0

a = 28; b = -17; c = -3;
Δ = b2-4ac
Δ = -172-4·28·(-3)
Δ = 625
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{625}=25$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-17)-25}{2*28}=\frac{-8}{56}
=-1/7
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-17)+25}{2*28}=\frac{42}{56}
=3/4
$