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h^2+18h+17=0
a = 1; b = 18; c = +17;
Δ = b2-4ac
Δ = 182-4·1·17
Δ = 256
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{256}=16$$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-16}{2*1}=\frac{-34}{2} =-17 $$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+16}{2*1}=\frac{-2}{2} =-1 $
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