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(12-2x)(8-2x)=40
We move all terms to the left:
(12-2x)(8-2x)-(40)=0
We add all the numbers together, and all the variables
(-2x+12)(-2x+8)-40=0
We multiply parentheses ..
(+4x^2-16x-24x+96)-40=0
We get rid of parentheses
4x^2-16x-24x+96-40=0
We add all the numbers together, and all the variables
4x^2-40x+56=0
a = 4; b = -40; c = +56;
Δ = b2-4ac
Δ = -402-4·4·56
Δ = 704
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{704}=\sqrt{64*11}=\sqrt{64}*\sqrt{11}=8\sqrt{11}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-8\sqrt{11}}{2*4}=\frac{40-8\sqrt{11}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+8\sqrt{11}}{2*4}=\frac{40+8\sqrt{11}}{8} $
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