216=(x+4)(4+4x)

Solution for 216=(x+4)(4+4x) equation:

216=(x+4)(4+4x)

We move all terms to the left:

216-((x+4)(4+4x))=0

We add all the numbers together, and all the variables

-((x+4)(4x+4))+216=0

We multiply parentheses ..

-((+4x^2+4x+16x+16))+216=0


We calculate terms in parentheses: -((+4x^2+4x+16x+16)), so:

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

We get rid of parentheses

4x^2+4x+16x+16

We add all the numbers together, and all the variables

4x^2+20x+16

Back to the equation:

-(4x^2+20x+16)


We get rid of parentheses

-4x^2-20x-16+216=0

We add all the numbers together, and all the variables

-4x^2-20x+200=0

a = -4; b = -20; c = +200;
Δ = b2-4ac
Δ = -202-4·(-4)·200
Δ = 3600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3600}=60$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-60}{2*-4}=\frac{-40}{-8}
=+5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+60}{2*-4}=\frac{80}{-8}
=-10
$