6x(2+1x)+20(x+2)=180

Solution for 6x(2+1x)+20(x+2)=180 equation:

6x(2+1x)+20(x+2)=180

We move all terms to the left:

6x(2+1x)+20(x+2)-(180)=0

We add all the numbers together, and all the variables

6x(x+2)+20(x+2)-180=0

We multiply parentheses

6x^2+12x+20x+40-180=0

We add all the numbers together, and all the variables

6x^2+32x-140=0

a = 6; b = 32; c = -140;
Δ = b2-4ac
Δ = 322-4·6·(-140)
Δ = 4384
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:

$\sqrt{\Delta}=\sqrt{4384}=\sqrt{16*274}=\sqrt{16}*\sqrt{274}=4\sqrt{274}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-4\sqrt{274}}{2*6}=\frac{-32-4\sqrt{274}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+4\sqrt{274}}{2*6}=\frac{-32+4\sqrt{274}}{12}
$