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6z-0.3z^2=0
a = -0.3; b = 6; c = 0;
Δ = b2-4ac
Δ = 62-4·(-0.3)·0
Δ = 36
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{36}=6$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-6}{2*-0.3}=\frac{-12}{-0.6} =+20 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6}{2*-0.3}=\frac{0}{-0.6} =0 $
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