(3x+8)(2x+4)=32

Solution for (3x+8)(2x+4)=32 equation:

(3x+8)(2x+4)=32

We move all terms to the left:

(3x+8)(2x+4)-(32)=0

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

6x^2+28x=0

a = 6; b = 28; c = 0;
Δ = b2-4ac
Δ = 282-4·6·0
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{-(28)-28}{2*6}=\frac{-56}{12}
=-4+2/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+28}{2*6}=\frac{0}{12}
=0
$