(2a)^2/(3-2a)(3-a)=2

Solution for (2a)^2/(3-2a)(3-a)=2 equation:

(2a)^2/(3-2a)(3-a)=2

We move all terms to the left:

(2a)^2/(3-2a)(3-a)-(2)=0


Domain of the equation: (3-2a)(3-a)!=0

We move all terms containing a to the left, all other terms to the right

-2a)(3-a!=-3

a∈R


We add all the numbers together, and all the variables

2a^2/(-2a+3)(-1a+3)-2=0

We multiply parentheses ..

2a^2/(+2a^2-6a-3a+9)-2=0

We multiply all the terms by the denominator


2a^2-2*(+2a^2-6a-3a+9)=0

We multiply parentheses

2a^2-4a^2+12a+6a-18=0

We add all the numbers together, and all the variables

-2a^2+18a-18=0

a = -2; b = 18; c = -18;
Δ = b2-4ac
Δ = 182-4·(-2)·(-18)
Δ = 180
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{180}=\sqrt{36*5}=\sqrt{36}*\sqrt{5}=6\sqrt{5}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-6\sqrt{5}}{2*-2}=\frac{-18-6\sqrt{5}}{-4}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+6\sqrt{5}}{2*-2}=\frac{-18+6\sqrt{5}}{-4}
$