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3n^2+24n+21=0
a = 3; b = 24; c = +21;
Δ = b2-4ac
Δ = 242-4·3·21
Δ = 324
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{324}=18$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-18}{2*3}=\frac{-42}{6} =-7 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+18}{2*3}=\frac{-6}{6} =-1 $
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