(3x-8)(x+12)=0

Solution for (3x-8)(x+12)=0 equation:

(3x-8)(x+12)=0

We multiply parentheses ..

(+3x^2+36x-8x-96)=0

We get rid of parentheses

3x^2+36x-8x-96=0

We add all the numbers together, and all the variables

3x^2+28x-96=0

a = 3; b = 28; c = -96;
Δ = b2-4ac
Δ = 282-4·3·(-96)
Δ = 1936
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1936}=44$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-44}{2*3}=\frac{-72}{6}
=-12
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+44}{2*3}=\frac{16}{6}
=2+2/3
$