3q2+9q+5=0

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Solution for 3q2+9q+5=0 equation: 



3q^2+9q+5=0

a = 3; b = 9; c = +5;
Δ = b2-4ac
Δ = 92-4·3·5
Δ = 21
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-\sqrt{21}}{2*3}=\frac{-9-\sqrt{21}}{6}
$


$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+\sqrt{21}}{2*3}=\frac{-9+\sqrt{21}}{6}
$

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