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1.5n^2+43n=0
a = 1.5; b = 43; c = 0;
Δ = b2-4ac
Δ = 432-4·1.5·0
Δ = 1849
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{1849}=43$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(43)-43}{2*1.5}=\frac{-86}{3} =-28+2/3 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(43)+43}{2*1.5}=\frac{0}{3} =0 $
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