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2n^2+5n+2=0
a = 2; b = 5; c = +2;
Δ = b2-4ac
Δ = 52-4·2·2
Δ = 9
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{9}=3$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-3}{2*2}=\frac{-8}{4} =-2 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+3}{2*2}=\frac{-2}{4} =-1/2 $
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