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49z^2-13z=0
a = 49; b = -13; c = 0;
Δ = b2-4ac
Δ = -132-4·49·0
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$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-13}{2*49}=\frac{0}{98} =0 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+13}{2*49}=\frac{26}{98} =13/49 $
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