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6z^2-13z=0
a = 6; b = -13; c = 0;
Δ = b2-4ac
Δ = -132-4·6·0
Δ = 169
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{169}=13$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-13}{2*6}=\frac{0}{12} =0 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+13}{2*6}=\frac{26}{12} =2+1/6 $
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