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(12+2x)(8+2x)=216
We move all terms to the left:
(12+2x)(8+2x)-(216)=0
We add all the numbers together, and all the variables
(2x+12)(2x+8)-216=0
We multiply parentheses ..
(+4x^2+16x+24x+96)-216=0
We get rid of parentheses
4x^2+16x+24x+96-216=0
We add all the numbers together, and all the variables
4x^2+40x-120=0
a = 4; b = 40; c = -120;
Δ = b2-4ac
Δ = 402-4·4·(-120)
Δ = 3520
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{3520}=\sqrt{64*55}=\sqrt{64}*\sqrt{55}=8\sqrt{55}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-8\sqrt{55}}{2*4}=\frac{-40-8\sqrt{55}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+8\sqrt{55}}{2*4}=\frac{-40+8\sqrt{55}}{8} $
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