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40z^2+27z=0
a = 40; b = 27; c = 0;
Δ = b2-4ac
Δ = 272-4·40·0
Δ = 729
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{729}=27$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(27)-27}{2*40}=\frac{-54}{80} =-27/40 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(27)+27}{2*40}=\frac{0}{80} =0 $
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