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42z^2-14z=0
a = 42; b = -14; c = 0;
Δ = b2-4ac
Δ = -142-4·42·0
Δ = 196
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{196}=14$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-14}{2*42}=\frac{0}{84} =0 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+14}{2*42}=\frac{28}{84} =1/3 $
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