2^2=x(6+x)

Solution for 2^2=x(6+x) equation:

2^2=x(6+x)

We move all terms to the left:

2^2-(x(6+x))=0

We add all the numbers together, and all the variables

-(x(x+6))+2^2=0

We add all the numbers together, and all the variables

-(x(x+6))+4=0


We calculate terms in parentheses: -(x(x+6)), so:

x(x+6)

We multiply parentheses

x^2+6x

Back to the equation:

-(x^2+6x)


We get rid of parentheses

-x^2-6x+4=0

We add all the numbers together, and all the variables

-1x^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}
$