9x+(x+6)*2-4(3-2x)*20x=20

Solution for 9x+(x+6)*2-4(3-2x)*20x=20 equation:

9x+(x+6)*2-4(3-2x)*20x=20

We move all terms to the left:

9x+(x+6)*2-4(3-2x)*20x-(20)=0

We add all the numbers together, and all the variables

9x+(x+6)*2-4(-2x+3)*20x-20=0

We multiply parentheses

160x^2+9x+2x-240x+12-20=0

We add all the numbers together, and all the variables

160x^2-229x-8=0

a = 160; b = -229; c = -8;
Δ = b2-4ac
Δ = -2292-4·160·(-8)
Δ = 57561
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{-(-229)-\sqrt{57561}}{2*160}=\frac{229-\sqrt{57561}}{320}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-229)+\sqrt{57561}}{2*160}=\frac{229+\sqrt{57561}}{320}
$