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4z^2-3z=0
a = 4; b = -3; c = 0;
Δ = b2-4ac
Δ = -32-4·4·0
Δ = 9
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{9}=3$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-3}{2*4}=\frac{0}{8} =0 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+3}{2*4}=\frac{6}{8} =3/4 $
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