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4z^2-6z=0
a = 4; b = -6; c = 0;
Δ = b2-4ac
Δ = -62-4·4·0
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$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-6}{2*4}=\frac{0}{8} =0 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+6}{2*4}=\frac{12}{8} =1+1/2 $
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