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28+y^2-16y=0
a = 1; b = -16; c = +28;
Δ = b2-4ac
Δ = -162-4·1·28
Δ = 144
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{144}=12$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-12}{2*1}=\frac{4}{2} =2 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+12}{2*1}=\frac{28}{2} =14 $
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