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147-3h^2=0
a = -3; b = 0; c = +147;
Δ = b2-4ac
Δ = 02-4·(-3)·147
Δ = 1764
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{1764}=42$$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-42}{2*-3}=\frac{-42}{-6} =+7 $$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+42}{2*-3}=\frac{42}{-6} =-7 $
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