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5y^2+19y=0
a = 5; b = 19; c = 0;
Δ = b2-4ac
Δ = 192-4·5·0
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{361}=19$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-19}{2*5}=\frac{-38}{10} =-3+4/5 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+19}{2*5}=\frac{0}{10} =0 $
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