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u^2-16u-17=0
a = 1; b = -16; c = -17;
Δ = b2-4ac
Δ = -162-4·1·(-17)
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{324}=18$$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-18}{2*1}=\frac{-2}{2} =-1 $$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+18}{2*1}=\frac{34}{2} =17 $
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