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(12-x)(20-x)=84
We move all terms to the left:
(12-x)(20-x)-(84)=0
We add all the numbers together, and all the variables
(-1x+12)(-1x+20)-84=0
We multiply parentheses ..
(+x^2-20x-12x+240)-84=0
We get rid of parentheses
x^2-20x-12x+240-84=0
We add all the numbers together, and all the variables
x^2-32x+156=0
a = 1; b = -32; c = +156;
Δ = b2-4ac
Δ = -322-4·1·156
Δ = 400
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{400}=20$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-20}{2*1}=\frac{12}{2} =6 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+20}{2*1}=\frac{52}{2} =26 $
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