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(8+2x)(10+2x)=168
We move all terms to the left:
(8+2x)(10+2x)-(168)=0
We add all the numbers together, and all the variables
(2x+8)(2x+10)-168=0
We multiply parentheses ..
(+4x^2+20x+16x+80)-168=0
We get rid of parentheses
4x^2+20x+16x+80-168=0
We add all the numbers together, and all the variables
4x^2+36x-88=0
a = 4; b = 36; c = -88;
Δ = b2-4ac
Δ = 362-4·4·(-88)
Δ = 2704
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{2704}=52$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(36)-52}{2*4}=\frac{-88}{8} =-11 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+52}{2*4}=\frac{16}{8} =2 $
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