(2)(x+2)(3x)=148

Solution for (2)(x+2)(3x)=148 equation:

(2)(x+2)(3x)=148

We move all terms to the left:

(2)(x+2)(3x)-(148)=0

We multiply parentheses

6x^2+12x-148=0

a = 6; b = 12; c = -148;
Δ = b2-4ac
Δ = 122-4·6·(-148)
Δ = 3696
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:

$\sqrt{\Delta}=\sqrt{3696}=\sqrt{16*231}=\sqrt{16}*\sqrt{231}=4\sqrt{231}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-4\sqrt{231}}{2*6}=\frac{-12-4\sqrt{231}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4\sqrt{231}}{2*6}=\frac{-12+4\sqrt{231}}{12}
$