(4x+4)(3x+2)=40

Solution for (4x+4)(3x+2)=40 equation:

(4x+4)(3x+2)=40

We move all terms to the left:

(4x+4)(3x+2)-(40)=0

We multiply parentheses ..

(+12x^2+8x+12x+8)-40=0

We get rid of parentheses

12x^2+8x+12x+8-40=0

We add all the numbers together, and all the variables

12x^2+20x-32=0

a = 12; b = 20; c = -32;
Δ = b2-4ac
Δ = 202-4·12·(-32)
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{-(20)-44}{2*12}=\frac{-64}{24}
=-2+2/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+44}{2*12}=\frac{24}{24}
=1
$