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(16+2x)(10+2x)=432
We move all terms to the left:
(16+2x)(10+2x)-(432)=0
We add all the numbers together, and all the variables
(2x+16)(2x+10)-432=0
We multiply parentheses ..
(+4x^2+20x+32x+160)-432=0
We get rid of parentheses
4x^2+20x+32x+160-432=0
We add all the numbers together, and all the variables
4x^2+52x-272=0
a = 4; b = 52; c = -272;
Δ = b2-4ac
Δ = 522-4·4·(-272)
Δ = 7056
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{7056}=84$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(52)-84}{2*4}=\frac{-136}{8} =-17 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(52)+84}{2*4}=\frac{32}{8} =4 $
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