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19x^2+25x=0
a = 19; b = 25; c = 0;
Δ = b2-4ac
Δ = 252-4·19·0
Δ = 625
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{625}=25$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-25}{2*19}=\frac{-50}{38} =-1+6/19 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+25}{2*19}=\frac{0}{38} =0 $
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