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42n^2+24n=0
a = 42; b = 24; c = 0;
Δ = b2-4ac
Δ = 242-4·42·0
Δ = 576
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{576}=24$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-24}{2*42}=\frac{-48}{84} =-4/7 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+24}{2*42}=\frac{0}{84} =0 $
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