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2x^2-41x=0
a = 2; b = -41; c = 0;
Δ = b2-4ac
Δ = -412-4·2·0
Δ = 1681
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{1681}=41$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-41)-41}{2*2}=\frac{0}{4} =0 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-41)+41}{2*2}=\frac{82}{4} =20+1/2 $
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