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h^2+23h+16=0
a = 1; b = 23; c = +16;
Δ = b2-4ac
Δ = 232-4·1·16
Δ = 465
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{-(23)-\sqrt{465}}{2*1}=\frac{-23-\sqrt{465}}{2} $$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+\sqrt{465}}{2*1}=\frac{-23+\sqrt{465}}{2} $
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