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(120-2x)(100-2x)=(2x)(x+10)
We move all terms to the left:
(120-2x)(100-2x)-((2x)(x+10))=0
We add all the numbers together, and all the variables
(-2x+120)(-2x+100)-(2x(x+10))=0
We multiply parentheses ..
(+4x^2-200x-240x+12000)-(2x(x+10))=0
We calculate terms in parentheses: -(2x(x+10)), so:We get rid of parentheses
2x(x+10)
We multiply parentheses
2x^2+20x
Back to the equation:
-(2x^2+20x)
4x^2-2x^2-200x-240x-20x+12000=0
We add all the numbers together, and all the variables
2x^2-460x+12000=0
a = 2; b = -460; c = +12000;
Δ = b2-4ac
Δ = -4602-4·2·12000
Δ = 115600
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{115600}=340$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-460)-340}{2*2}=\frac{120}{4} =30 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-460)+340}{2*2}=\frac{800}{4} =200 $
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