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8y^2+28y=0
a = 8; b = 28; c = 0;
Δ = b2-4ac
Δ = 282-4·8·0
Δ = 784
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{784}=28$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-28}{2*8}=\frac{-56}{16} =-3+1/2 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+28}{2*8}=\frac{0}{16} =0 $
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