(x-4)(3x+2)=-16

Solution for (x-4)(3x+2)=-16 equation:

(x-4)(3x+2)=-16

We move all terms to the left:

(x-4)(3x+2)-(-16)=0

We add all the numbers together, and all the variables

(x-4)(3x+2)+16=0

We multiply parentheses ..

(+3x^2+2x-12x-8)+16=0

We get rid of parentheses

3x^2+2x-12x-8+16=0

We add all the numbers together, and all the variables

3x^2-10x+8=0

a = 3; b = -10; c = +8;
Δ = b2-4ac
Δ = -102-4·3·8
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2}{2*3}=\frac{8}{6}
=1+1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2}{2*3}=\frac{12}{6}
=2
$