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16y^2+64y=0
a = 16; b = 64; c = 0;
Δ = b2-4ac
Δ = 642-4·16·0
Δ = 4096
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{4096}=64$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(64)-64}{2*16}=\frac{-128}{32} =-4 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(64)+64}{2*16}=\frac{0}{32} =0 $
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