(x+8)x=(2x+1)(2x+1)

Solution for (x+8)x=(2x+1)(2x+1) equation:

(x+8)x=(2x+1)(2x+1)

We move all terms to the left:

(x+8)x-((2x+1)(2x+1))=0

We multiply parentheses

x^2+8x-((2x+1)(2x+1))=0

We multiply parentheses ..

x^2-((+4x^2+2x+2x+1))+8x=0


We calculate terms in parentheses: -((+4x^2+2x+2x+1)), so:

(+4x^2+2x+2x+1)

We get rid of parentheses

4x^2+2x+2x+1

We add all the numbers together, and all the variables

4x^2+4x+1

Back to the equation:

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


We add all the numbers together, and all the variables

x^2+8x-(4x^2+4x+1)=0

We get rid of parentheses

x^2-4x^2+8x-4x-1=0

We add all the numbers together, and all the variables

-3x^2+4x-1=0

a = -3; b = 4; c = -1;
Δ = b2-4ac
Δ = 42-4·(-3)·(-1)
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2}{2*-3}=\frac{-6}{-6}
=1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2}{2*-3}=\frac{-2}{-6}
=1/3
$