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12z^2+3z=0
a = 12; b = 3; c = 0;
Δ = b2-4ac
Δ = 32-4·12·0
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$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3}{2*12}=\frac{-6}{24} =-1/4 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3}{2*12}=\frac{0}{24} =0 $
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