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(16+2x)(12+2x)+192=396
We move all terms to the left:
(16+2x)(12+2x)+192-(396)=0
We add all the numbers together, and all the variables
(2x+16)(2x+12)+192-396=0
We add all the numbers together, and all the variables
(2x+16)(2x+12)-204=0
We multiply parentheses ..
(+4x^2+24x+32x+192)-204=0
We get rid of parentheses
4x^2+24x+32x+192-204=0
We add all the numbers together, and all the variables
4x^2+56x-12=0
a = 4; b = 56; c = -12;
Δ = b2-4ac
Δ = 562-4·4·(-12)
Δ = 3328
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{3328}=\sqrt{256*13}=\sqrt{256}*\sqrt{13}=16\sqrt{13}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(56)-16\sqrt{13}}{2*4}=\frac{-56-16\sqrt{13}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(56)+16\sqrt{13}}{2*4}=\frac{-56+16\sqrt{13}}{8} $
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