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50z^2+13z=0
a = 50; b = 13; c = 0;
Δ = b2-4ac
Δ = 132-4·50·0
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$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-13}{2*50}=\frac{-26}{100} =-13/50 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+13}{2*50}=\frac{0}{100} =0 $
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