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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(+6x^2-8x+12x-16)=0

We get rid of parentheses

6x^2-8x+12x-16=0

We add all the numbers together, and all the variables

6x^2+4x-16=0

a = 6; b = 4; c = -16;
Δ = b2-4ac
Δ = 42-4·6·(-16)
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*6}=\frac{-24}{12}
=-2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+20}{2*6}=\frac{16}{12}
=1+1/3
$