(3x-8)(x+4)=0

Solution for (3x-8)(x+4)=0 equation:

(3x-8)(x+4)=0

We multiply parentheses ..

(+3x^2+12x-8x-32)=0

We get rid of parentheses

3x^2+12x-8x-32=0

We add all the numbers together, and all the variables

3x^2+4x-32=0

a = 3; b = 4; c = -32;
Δ = b2-4ac
Δ = 42-4·3·(-32)
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-20}{2*3}=\frac{-24}{6}
=-4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+20}{2*3}=\frac{16}{6}
=2+2/3
$