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13y^2+42y=0
a = 13; b = 42; c = 0;
Δ = b2-4ac
Δ = 422-4·13·0
Δ = 1764
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{1764}=42$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-42}{2*13}=\frac{-84}{26} =-3+3/13 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+42}{2*13}=\frac{0}{26} =0 $
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