v^2+9v+10=0

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Solution for v^2+9v+10=0 equation: 



v^2+9v+10=0

a = 1; b = 9; c = +10;
Δ = b2-4ac
Δ = 92-4·1·10
Δ = 41
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-\sqrt{41}}{2*1}=\frac{-9-\sqrt{41}}{2}
$


$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+\sqrt{41}}{2*1}=\frac{-9+\sqrt{41}}{2}
$

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