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h^2+17h-25=0
a = 1; b = 17; c = -25;
Δ = b2-4ac
Δ = 172-4·1·(-25)
Δ = 389
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}$$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-\sqrt{389}}{2*1}=\frac{-17-\sqrt{389}}{2} $$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+\sqrt{389}}{2*1}=\frac{-17+\sqrt{389}}{2} $
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