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^3-2Z^2+5Z=0
We add all the numbers together, and all the variables
-2Z^2+5Z=0
a = -2; b = 5; c = 0;
Δ = b2-4ac
Δ = 52-4·(-2)·0
Δ = 25
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{25}=5$$Z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-5}{2*-2}=\frac{-10}{-4} =2+1/2 $$Z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+5}{2*-2}=\frac{0}{-4} =0 $
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