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10y^2+10y=0
a = 10; b = 10; c = 0;
Δ = b2-4ac
Δ = 102-4·10·0
Δ = 100
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{100}=10$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-10}{2*10}=\frac{-20}{20} =-1 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+10}{2*10}=\frac{0}{20} =0 $
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