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

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

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

We multiply parentheses ..

(+12x^2-16x+3x-4)=0

We get rid of parentheses

12x^2-16x+3x-4=0

We add all the numbers together, and all the variables

12x^2-13x-4=0

a = 12; b = -13; c = -4;
Δ = b2-4ac
Δ = -132-4·12·(-4)
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-19}{2*12}=\frac{-6}{24}
=-1/4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+19}{2*12}=\frac{32}{24}
=1+1/3
$