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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

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


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

(x^2-4x)

We get rid of parentheses

x^2-4x

Back to the equation:

-(x^2-4x)


We get rid of parentheses

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

We add all the numbers together, and all the variables

-1x^2+6x+16=0

a = -1; b = 6; c = +16;
Δ = b2-4ac
Δ = 62-4·(-1)·16
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-10}{2*-1}=\frac{-16}{-2}
=+8
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+10}{2*-1}=\frac{4}{-2}
=-2
$