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19d^2-39d=0
a = 19; b = -39; c = 0;
Δ = b2-4ac
Δ = -392-4·19·0
Δ = 1521
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{1521}=39$$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-39)-39}{2*19}=\frac{0}{38} =0 $$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-39)+39}{2*19}=\frac{78}{38} =2+1/19 $
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