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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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


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

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

We get rid of parentheses

-3x^2+1x-12x+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{-24}{6}
=-4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+13}{2*3}=\frac{2}{6}
=1/3
$