x*x+(x+3)(x+3)=(x+4)(x+4)

Solution for x*x+(x+3)(x+3)=(x+4)(x+4) equation:

x*x+(x+3)(x+3)=(x+4)(x+4)

We move all terms to the left:

x*x+(x+3)(x+3)-((x+4)(x+4))=0

Wy multiply elements

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

We multiply parentheses ..

x^2+(+x^2+3x+3x+9)-((x+4)(x+4))=0


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

(x+4)(x+4)

We multiply parentheses ..

(+x^2+4x+4x+16)

We get rid of parentheses

x^2+4x+4x+16

We add all the numbers together, and all the variables

x^2+8x+16

Back to the equation:

-(x^2+8x+16)


We get rid of parentheses

x^2+x^2-x^2+3x+3x-8x+9-16=0

We add all the numbers together, and all the variables

x^2-2x-7=0

a = 1; b = -2; c = -7;
Δ = b2-4ac
Δ = -22-4·1·(-7)
Δ = 32
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:

$\sqrt{\Delta}=\sqrt{32}=\sqrt{16*2}=\sqrt{16}*\sqrt{2}=4\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-4\sqrt{2}}{2*1}=\frac{2-4\sqrt{2}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+4\sqrt{2}}{2*1}=\frac{2+4\sqrt{2}}{2}
$