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2x-x^2=0.1
We move all terms to the left:
2x-x^2-(0.1)=0
We add all the numbers together, and all the variables
-1x^2+2x-0.1=0
a = -1; b = 2; c = -0.1;
Δ = b2-4ac
Δ = 22-4·(-1)·(-0.1)
Δ = 3.6
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{-(2)-\sqrt{3.6}}{2*-1}=\frac{-2-\sqrt{3.6}}{-2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+\sqrt{3.6}}{2*-1}=\frac{-2+\sqrt{3.6}}{-2} $
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