132=192-(16+2x)(12+2x)

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Solution for 132=192-(16+2x)(12+2x) equation:



132=192-(16+2x)(12+2x)
We move all terms to the left:
132-(192-(16+2x)(12+2x))=0
We add all the numbers together, and all the variables
-(192-(2x+16)(2x+12))+132=0
We multiply parentheses ..
-(192-(+4x^2+24x+32x+192))+132=0
We calculate terms in parentheses: -(192-(+4x^2+24x+32x+192)), so:
192-(+4x^2+24x+32x+192)
determiningTheFunctionDomain -(+4x^2+24x+32x+192)+192
We get rid of parentheses
-4x^2-24x-32x-192+192
We add all the numbers together, and all the variables
-4x^2-56x
Back to the equation:
-(-4x^2-56x)
We get rid of parentheses
4x^2+56x+132=0
a = 4; b = 56; c = +132;
Δ = b2-4ac
Δ = 562-4·4·132
Δ = 1024
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1024}=32$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(56)-32}{2*4}=\frac{-88}{8} =-11 $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(56)+32}{2*4}=\frac{-24}{8} =-3 $

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