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

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

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

We move all terms to the left:

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

We multiply parentheses

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

We multiply parentheses ..

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


We calculate terms in parentheses: -((+x^2-2x-2x+4)), so:

(+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 add all the numbers together, and all the variables

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

x^2+6x-4=0

a = 1; b = 6; c = -4;
Δ = b2-4ac
Δ = 62-4·1·(-4)
Δ = 52
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{52}=\sqrt{4*13}=\sqrt{4}*\sqrt{13}=2\sqrt{13}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{13}}{2*1}=\frac{-6-2\sqrt{13}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{13}}{2*1}=\frac{-6+2\sqrt{13}}{2}
$