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(2y^2)-40y+128=0
a = 2; b = -40; c = +128;
Δ = b2-4ac
Δ = -402-4·2·128
Δ = 576
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{576}=24$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-24}{2*2}=\frac{16}{4} =4 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+24}{2*2}=\frac{64}{4} =16 $
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