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(12+2x)(8+2x)=140
We move all terms to the left:
(12+2x)(8+2x)-(140)=0
We add all the numbers together, and all the variables
(2x+12)(2x+8)-140=0
We multiply parentheses ..
(+4x^2+16x+24x+96)-140=0
We get rid of parentheses
4x^2+16x+24x+96-140=0
We add all the numbers together, and all the variables
4x^2+40x-44=0
a = 4; b = 40; c = -44;
Δ = b2-4ac
Δ = 402-4·4·(-44)
Δ = 2304
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{2304}=48$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-48}{2*4}=\frac{-88}{8} =-11 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+48}{2*4}=\frac{8}{8} =1 $
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