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

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



216=(x+4)(4x+4)
We move all terms to the left:
216-((x+4)(4x+4))=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 $

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