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

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

(x-2)(2x-3)=x^2

We move all terms to the left:

(x-2)(2x-3)-(x^2)=0

determiningTheFunctionDomain
-x^2+(x-2)(2x-3)=0

We add all the numbers together, and all the variables

-1x^2+(x-2)(2x-3)=0

We multiply parentheses ..

-1x^2+(+2x^2-3x-4x+6)=0

We get rid of parentheses

-1x^2+2x^2-3x-4x+6=0

We add all the numbers together, and all the variables

x^2-7x+6=0

a = 1; b = -7; c = +6;
Δ = b2-4ac
Δ = -72-4·1·6
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-5}{2*1}=\frac{2}{2}
=1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+5}{2*1}=\frac{12}{2}
=6
$