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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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


We calculate terms in parentheses: -(8(2x+1)(-1x+2)), so:

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

We multiply parentheses ..

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

We multiply parentheses

-16x^2+32x-8x+16

We add all the numbers together, and all the variables

-16x^2+24x+16

Back to the equation:

-(-16x^2+24x+16)


We get rid of parentheses

4x^2+16x^2+2x+2x-24x+1-16=0

We add all the numbers together, and all the variables

20x^2-20x-15=0

a = 20; b = -20; c = -15;
Δ = b2-4ac
Δ = -202-4·20·(-15)
Δ = 1600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1600}=40$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-40}{2*20}=\frac{-20}{40}
=-1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+40}{2*20}=\frac{60}{40}
=1+1/2
$