(x)(3x+2)=165

Solution for (x)(3x+2)=165 equation:

(x)(3x+2)=165

We move all terms to the left:

(x)(3x+2)-(165)=0

We multiply parentheses

3x^2+2x-165=0

a = 3; b = 2; c = -165;
Δ = b2-4ac
Δ = 22-4·3·(-165)
Δ = 1984
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:

$\sqrt{\Delta}=\sqrt{1984}=\sqrt{64*31}=\sqrt{64}*\sqrt{31}=8\sqrt{31}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-8\sqrt{31}}{2*3}=\frac{-2-8\sqrt{31}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+8\sqrt{31}}{2*3}=\frac{-2+8\sqrt{31}}{6}
$