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29y^2+45y=0
a = 29; b = 45; c = 0;
Δ = b2-4ac
Δ = 452-4·29·0
Δ = 2025
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{2025}=45$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(45)-45}{2*29}=\frac{-90}{58} =-1+16/29 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(45)+45}{2*29}=\frac{0}{58} =0 $
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