2n2+9n+10=0

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Solution for 2n2+9n+10=0 equation: 



2n^2+9n+10=0

a = 2; b = 9; c = +10;
Δ = b2-4ac
Δ = 92-4·2·10
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$


$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-1}{2*2}=\frac{-10}{4}
=-2+1/2
$


$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+1}{2*2}=\frac{-8}{4}
=-2
$

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