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2h^2+10h-72=0
a = 2; b = 10; c = -72;
Δ = b2-4ac
Δ = 102-4·2·(-72)
Δ = 676
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{676}=26$$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-26}{2*2}=\frac{-36}{4} =-9 $$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+26}{2*2}=\frac{16}{4} =4 $
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