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13d^2-2d=0
a = 13; b = -2; c = 0;
Δ = b2-4ac
Δ = -22-4·13·0
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{4}=2$$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2}{2*13}=\frac{0}{26} =0 $$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2}{2*13}=\frac{4}{26} =2/13 $
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