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

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

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

We move all terms to the left:

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


We calculate terms in parentheses: -(2x(3x-10)+50), so:

2x(3x-10)+50

We multiply parentheses

6x^2-20x+50

Back to the equation:

-(6x^2-20x+50)


We get rid of parentheses

-6x^2+20x-50+180=0

We add all the numbers together, and all the variables

-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}
$