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14x^2+42x=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:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{1764}=42$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-42}{2*14}=\frac{-84}{28} =-3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+42}{2*14}=\frac{0}{28} =0 $
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