50+(2x)(3x-10)=180

Solution for 50+(2x)(3x-10)=180 equation:

50+(2x)(3x-10)=180

We move all terms to the left:

50+(2x)(3x-10)-(180)=0

We add all the numbers together, and all the variables

2x(3x-10)-130=0

We multiply parentheses

6x^2-20x-130=0

a = 6; b = -20; c = -130;
Δ = b2-4ac
Δ = -202-4·6·(-130)
Δ = 3520
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:

$\sqrt{\Delta}=\sqrt{3520}=\sqrt{64*55}=\sqrt{64}*\sqrt{55}=8\sqrt{55}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-8\sqrt{55}}{2*6}=\frac{20-8\sqrt{55}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+8\sqrt{55}}{2*6}=\frac{20+8\sqrt{55}}{12}
$