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50x+400=100x-x^2
We move all terms to the left:
50x+400-(100x-x^2)=0
We get rid of parentheses
x^2-100x+50x+400=0
We add all the numbers together, and all the variables
x^2-50x+400=0
a = 1; b = -50; c = +400;
Δ = b2-4ac
Δ = -502-4·1·400
Δ = 900
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{900}=30$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-50)-30}{2*1}=\frac{20}{2} =10 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-50)+30}{2*1}=\frac{80}{2} =40 $
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