(X-2)(x-2)=3(2x-3)(2-x)

Solution for (X-2)(x-2)=3(2x-3)(2-x) equation:

(X-2)(X-2)=3(2X-3)(2-X)

We move all terms to the left:

(X-2)(X-2)-(3(2X-3)(2-X))=0

We add all the numbers together, and all the variables

(X-2)(X-2)-(3(2X-3)(-1X+2))=0

We multiply parentheses ..

(+X^2-2X-2X+4)-(3(2X-3)(-1X+2))=0


We calculate terms in parentheses: -(3(2X-3)(-1X+2)), so:

3(2X-3)(-1X+2)

We multiply parentheses ..

3(-2X^2+4X+3X-6)

We multiply parentheses

-6X^2+12X+9X-18

We add all the numbers together, and all the variables

-6X^2+21X-18

Back to the equation:

-(-6X^2+21X-18)


We get rid of parentheses

X^2+6X^2-2X-2X-21X+4+18=0

We add all the numbers together, and all the variables

7X^2-25X+22=0

a = 7; b = -25; c = +22;
Δ = b2-4ac
Δ = -252-4·7·22
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-25)-3}{2*7}=\frac{22}{14}
=1+4/7
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+3}{2*7}=\frac{28}{14}
=2
$