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36z^2+64z=0
a = 36; b = 64; c = 0;
Δ = b2-4ac
Δ = 642-4·36·0
Δ = 4096
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{4096}=64$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(64)-64}{2*36}=\frac{-128}{72} =-1+7/9 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(64)+64}{2*36}=\frac{0}{72} =0 $
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