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14z^2-42z=0
a = 14; b = -42; c = 0;
Δ = b2-4ac
Δ = -422-4·14·0
Δ = 1764
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{1764}=42$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-42)-42}{2*14}=\frac{0}{28} =0 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-42)+42}{2*14}=\frac{84}{28} =3 $
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