(2x+1)(2x+1)=(x-2)(x-2)

Solution for (2x+1)(2x+1)=(x-2)(x-2) equation:

(2x+1)(2x+1)=(x-2)(x-2)

We move all terms to the left:

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

We multiply parentheses ..

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


We calculate terms in parentheses: -((x-2)(x-2)), so:

(x-2)(x-2)

We multiply parentheses ..

(+x^2-2x-2x+4)

We get rid of parentheses

x^2-2x-2x+4

We add all the numbers together, and all the variables

x^2-4x+4

Back to the equation:

-(x^2-4x+4)


We get rid of parentheses

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

We add all the numbers together, and all the variables

3x^2+8x-3=0

a = 3; b = 8; c = -3;
Δ = b2-4ac
Δ = 82-4·3·(-3)
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-10}{2*3}=\frac{-18}{6}
=-3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+10}{2*3}=\frac{2}{6}
=1/3
$