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28y^2+21y=0
a = 28; b = 21; c = 0;
Δ = b2-4ac
Δ = 212-4·28·0
Δ = 441
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{441}=21$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-21}{2*28}=\frac{-42}{56} =-3/4 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+21}{2*28}=\frac{0}{56} =0 $
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