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y^2-16y-512=0
a = 1; b = -16; c = -512;
Δ = b2-4ac
Δ = -162-4·1·(-512)
Δ = 2304
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{2304}=48$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-48}{2*1}=\frac{-32}{2} =-16 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+48}{2*1}=\frac{64}{2} =32 $
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