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10x^2=20x=0
We move all terms to the left:
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-20}{2*10}=\frac{0}{20} =0 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+20}{2*10}=\frac{40}{20} =2 $
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