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(h^2)-6h-14=0
a = 1; b = -6; c = -14;
Δ = b2-4ac
Δ = -62-4·1·(-14)
Δ = 92
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}$$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{92}=\sqrt{4*23}=\sqrt{4}*\sqrt{23}=2\sqrt{23}$$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{23}}{2*1}=\frac{6-2\sqrt{23}}{2} $$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{23}}{2*1}=\frac{6+2\sqrt{23}}{2} $
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