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13z^2+17z=0
a = 13; b = 17; c = 0;
Δ = b2-4ac
Δ = 172-4·13·0
Δ = 289
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{289}=17$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-17}{2*13}=\frac{-34}{26} =-1+4/13 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+17}{2*13}=\frac{0}{26} =0 $
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