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2n^2+21n+10=0
a = 2; b = 21; c = +10;
Δ = b2-4ac
Δ = 212-4·2·10
Δ = 361
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{361}=19$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-19}{2*2}=\frac{-40}{4} =-10 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+19}{2*2}=\frac{-2}{4} =-1/2 $
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