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4z(9z-4)=0
We multiply parentheses
36z^2-16z=0
a = 36; b = -16; c = 0;
Δ = b2-4ac
Δ = -162-4·36·0
Δ = 256
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{256}=16$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-16}{2*36}=\frac{0}{72} =0 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+16}{2*36}=\frac{32}{72} =4/9 $
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