x(x+4)=(x+1)2

Solution for x(x+4)=(x+1)2 equation:

x(x+4)=(x+1)2

We move all terms to the left:

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

We multiply parentheses

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


We calculate terms in parentheses: -((x+1)2), so:

(x+1)2

We multiply parentheses

2x+2

Back to the equation:

-(2x+2)


We get rid of parentheses

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

We add all the numbers together, and all the variables

x^2+2x-2=0

a = 1; b = 2; c = -2;
Δ = b2-4ac
Δ = 22-4·1·(-2)
Δ = 12
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:

$\sqrt{\Delta}=\sqrt{12}=\sqrt{4*3}=\sqrt{4}*\sqrt{3}=2\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{3}}{2*1}=\frac{-2-2\sqrt{3}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{3}}{2*1}=\frac{-2+2\sqrt{3}}{2}
$