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20x=10x^2
We move all terms to the left:
20x-(10x^2)=0
determiningTheFunctionDomain -10x^2+20x=0
a = -10; b = 20; c = 0;
Δ = b2-4ac
Δ = 202-4·(-10)·0
Δ = 400
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{400}=20$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-20}{2*-10}=\frac{-40}{-20} =+2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+20}{2*-10}=\frac{0}{-20} =0 $
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