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h^2+16h-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:$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{324}=18$$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-18}{2*1}=\frac{-34}{2} =-17 $$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+18}{2*1}=\frac{2}{2} =1 $
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