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h^2+3h-140=0
a = 1; b = 3; c = -140;
Δ = b2-4ac
Δ = 32-4·1·(-140)
Δ = 569
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{-(3)-\sqrt{569}}{2*1}=\frac{-3-\sqrt{569}}{2} $$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{569}}{2*1}=\frac{-3+\sqrt{569}}{2} $
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