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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

-((+3x^2+12x-1x-4))=0


We calculate terms in parentheses: -((+3x^2+12x-1x-4)), so:

(+3x^2+12x-1x-4)

We get rid of parentheses

3x^2+12x-1x-4

We add all the numbers together, and all the variables

3x^2+11x-4

Back to the equation:

-(3x^2+11x-4)


We get rid of parentheses

-3x^2-11x+4=0

a = -3; b = -11; c = +4;
Δ = b2-4ac
Δ = -112-4·(-3)·4
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-13}{2*-3}=\frac{-2}{-6}
=1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+13}{2*-3}=\frac{24}{-6}
=-4
$