(3x-2)(x+4)=48

Solution for (3x-2)(x+4)=48 equation:

(3x-2)(x+4)=48

We move all terms to the left:

(3x-2)(x+4)-(48)=0

We multiply parentheses ..

(+3x^2+12x-2x-8)-48=0

We get rid of parentheses

3x^2+12x-2x-8-48=0

We add all the numbers together, and all the variables

3x^2+10x-56=0

a = 3; b = 10; c = -56;
Δ = b2-4ac
Δ = 102-4·3·(-56)
Δ = 772
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{772}=\sqrt{4*193}=\sqrt{4}*\sqrt{193}=2\sqrt{193}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{193}}{2*3}=\frac{-10-2\sqrt{193}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{193}}{2*3}=\frac{-10+2\sqrt{193}}{6}
$