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h^2+5h-295=0
a = 1; b = 5; c = -295;
Δ = b2-4ac
Δ = 52-4·1·(-295)
Δ = 1205
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{-(5)-\sqrt{1205}}{2*1}=\frac{-5-\sqrt{1205}}{2} $$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{1205}}{2*1}=\frac{-5+\sqrt{1205}}{2} $
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