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

Solution for 16x^2+4x=(4x+2)(4x+2x) equation:

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

We move all terms to the left:

16x^2+4x-((4x+2)(4x+2x))=0

We add all the numbers together, and all the variables

16x^2+4x-((4x+2)(+6x))=0

We multiply parentheses ..

16x^2-((+24x^2+12x))+4x=0


We calculate terms in parentheses: -((+24x^2+12x)), so:

(+24x^2+12x)

We get rid of parentheses

24x^2+12x

Back to the equation:

-(24x^2+12x)


We add all the numbers together, and all the variables

16x^2+4x-(24x^2+12x)=0

We get rid of parentheses

16x^2-24x^2+4x-12x=0

We add all the numbers together, and all the variables

-8x^2-8x=0

a = -8; b = -8; c = 0;
Δ = b2-4ac
Δ = -82-4·(-8)·0
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-8}{2*-8}=\frac{0}{-16}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+8}{2*-8}=\frac{16}{-16}
=-1
$