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25y^2+22y=0
a = 25; b = 22; c = 0;
Δ = b2-4ac
Δ = 222-4·25·0
Δ = 484
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{484}=22$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-22}{2*25}=\frac{-44}{50} =-22/25 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+22}{2*25}=\frac{0}{50} =0 $
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