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32y-2y^2=0
a = -2; b = 32; c = 0;
Δ = b2-4ac
Δ = 322-4·(-2)·0
Δ = 1024
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{1024}=32$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-32}{2*-2}=\frac{-64}{-4} =+16 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+32}{2*-2}=\frac{0}{-4} =0 $
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