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21n+12n^2=0
a = 12; b = 21; c = 0;
Δ = b2-4ac
Δ = 212-4·12·0
Δ = 441
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{441}=21$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-21}{2*12}=\frac{-42}{24} =-1+3/4 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+21}{2*12}=\frac{0}{24} =0 $
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