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(20-x)(23-x)=120
We move all terms to the left:
(20-x)(23-x)-(120)=0
We add all the numbers together, and all the variables
(-1x+20)(-1x+23)-120=0
We multiply parentheses ..
(+x^2-23x-20x+460)-120=0
We get rid of parentheses
x^2-23x-20x+460-120=0
We add all the numbers together, and all the variables
x^2-43x+340=0
a = 1; b = -43; c = +340;
Δ = b2-4ac
Δ = -432-4·1·340
Δ = 489
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-43)-\sqrt{489}}{2*1}=\frac{43-\sqrt{489}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-43)+\sqrt{489}}{2*1}=\frac{43+\sqrt{489}}{2} $
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