2x(3x+2)=162

Solution for 2x(3x+2)=162 equation:

2x(3x+2)=162

We move all terms to the left:

2x(3x+2)-(162)=0

We multiply parentheses

6x^2+4x-162=0

a = 6; b = 4; c = -162;
Δ = b2-4ac
Δ = 42-4·6·(-162)
Δ = 3904
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:

$\sqrt{\Delta}=\sqrt{3904}=\sqrt{64*61}=\sqrt{64}*\sqrt{61}=8\sqrt{61}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-8\sqrt{61}}{2*6}=\frac{-4-8\sqrt{61}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+8\sqrt{61}}{2*6}=\frac{-4+8\sqrt{61}}{12}
$