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

Solution for (2x+1)(2x+1)=(x-3)(x-3) equation:

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

We move all terms to the left:

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

We multiply parentheses ..

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


We calculate terms in parentheses: -((x-3)(x-3)), so:

(x-3)(x-3)

We multiply parentheses ..

(+x^2-3x-3x+9)

We get rid of parentheses

x^2-3x-3x+9

We add all the numbers together, and all the variables

x^2-6x+9

Back to the equation:

-(x^2-6x+9)


We get rid of parentheses

4x^2-x^2+2x+2x+6x+1-9=0

We add all the numbers together, and all the variables

3x^2+10x-8=0

a = 3; b = 10; c = -8;
Δ = b2-4ac
Δ = 102-4·3·(-8)
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-14}{2*3}=\frac{-24}{6}
=-4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+14}{2*3}=\frac{4}{6}
=2/3
$