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h^2-16h-80=0
a = 1; b = -16; c = -80;
Δ = b2-4ac
Δ = -162-4·1·(-80)
Δ = 576
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{576}=24$$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-24}{2*1}=\frac{-8}{2} =-4 $$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+24}{2*1}=\frac{40}{2} =20 $
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