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13z^2-41z=0
a = 13; b = -41; c = 0;
Δ = b2-4ac
Δ = -412-4·13·0
Δ = 1681
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{1681}=41$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-41)-41}{2*13}=\frac{0}{26} =0 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-41)+41}{2*13}=\frac{82}{26} =3+2/13 $
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