Square of sum: Resulting in a quadratic equation

To solve the equation, let's start by getting rid of the denominators.

To do this, we'll multiply the denominators:

$(2x+1)^2\cdot(2x+1)+(x+2)^2\cdot(x+2)=4.5x(2x+1)(x+2)$

Let's start by opening the parentheses on the left side, mainly using the distributive property:

$(4x^2+4x+1)\cdot(2x+1)+(x^2+4x+4)\cdot(x+2)=4.5x(2x+1)(x+2)$

Let's continue by opening the parentheses on the right side of the equation:

$(4x^2+4x+1)\cdot(2x+1)+(x^2+4x+4)\cdot(x+2)=4.5x(2x^2+5x+2)$Let's continue and open the parentheses on the right side of the equation:

$(4x^2+4x+1)\cdot(2x+1)+(x^2+4x+4)\cdot(x+2)=9x^3+22.5x+9x$

Now let's go back and simplify the parentheses on the left side of the equation:

$8x^3+8x^2+2x+4x^2+4x+1+x^3+4x^2+4x+2x^2+8x+8=9x^3+22.5x+9x$

Let's combine like terms:

$9x^3+18x^2+18x+9=9x^3+22.5x+9x$

Notice that all terms can be divided by 9, so let's do that:

$x^3+2x^2+2x+1=x^3+2.5x+x$

Let's move all numbers to one side:

$x^3-x^3+2x^2-2.5x^2+2x-x+9=0$

And we get:

$0.5x^2-x-1=0$

To get rid of the one-half coefficient, let's multiply the entire equation by 2

$x^2-2x-2=0$

Now we can use the square root formula, and we get-

$\frac{2±\sqrt{12}}{2}$

Let's use the properties of square roots to simplify the square root of 12:

$\frac{2±2\sqrt{3}}{2}$Let's divide both numerator and denominator by 2 and we get:

$1±\sqrt{3}$