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19q^2+39q=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:$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{1521}=39$$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(39)-39}{2*19}=\frac{-78}{38} =-2+1/19 $$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(39)+39}{2*19}=\frac{0}{38} =0 $
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