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

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

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

3x^2-8x-16=0

a = 3; b = -8; c = -16;
Δ = b2-4ac
Δ = -82-4·3·(-16)
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-16}{2*3}=\frac{-8}{6}
=-1+1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+16}{2*3}=\frac{24}{6}
=4
$