3x(2x+11)+(4x-2)=180

Solution for 3x(2x+11)+(4x-2)=180 equation:

3x(2x+11)+(4x-2)=180

We move all terms to the left:

3x(2x+11)+(4x-2)-(180)=0

We multiply parentheses

6x^2+33x+(4x-2)-180=0

We get rid of parentheses

6x^2+33x+4x-2-180=0

We add all the numbers together, and all the variables

6x^2+37x-182=0

a = 6; b = 37; c = -182;
Δ = b2-4ac
Δ = 372-4·6·(-182)
Δ = 5737
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(37)-\sqrt{5737}}{2*6}=\frac{-37-\sqrt{5737}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(37)+\sqrt{5737}}{2*6}=\frac{-37+\sqrt{5737}}{12}
$