30=3x(20-x)*2

Solution for 30=3x(20-x)*2 equation:

30=3x(20-x)*2

We move all terms to the left:

30-(3x(20-x)*2)=0

We add all the numbers together, and all the variables

-(3x(-1x+20)*2)+30=0


We calculate terms in parentheses: -(3x(-1x+20)*2), so:

3x(-1x+20)*2

We multiply parentheses

-6x^2+120x

Back to the equation:

-(-6x^2+120x)


We get rid of parentheses

6x^2-120x+30=0

a = 6; b = -120; c = +30;
Δ = b2-4ac
Δ = -1202-4·6·30
Δ = 13680
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:

$\sqrt{\Delta}=\sqrt{13680}=\sqrt{144*95}=\sqrt{144}*\sqrt{95}=12\sqrt{95}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-120)-12\sqrt{95}}{2*6}=\frac{120-12\sqrt{95}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-120)+12\sqrt{95}}{2*6}=\frac{120+12\sqrt{95}}{12}
$