(3x+8)(2x-4)=0

Solution for (3x+8)(2x-4)=0 equation:

(3x+8)(2x-4)=0

We multiply parentheses ..

(+6x^2-12x+16x-32)=0

We get rid of parentheses

6x^2-12x+16x-32=0

We add all the numbers together, and all the variables

6x^2+4x-32=0

a = 6; b = 4; c = -32;
Δ = b2-4ac
Δ = 42-4·6·(-32)
Δ = 784
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{784}=28$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-28}{2*6}=\frac{-32}{12}
=-2+2/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+28}{2*6}=\frac{24}{12}
=2
$