3q2+10q+3=0

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Solution for 3q2+10q+3=0 equation:



3q^2+10q+3=0
a = 3; b = 10; c = +3;
Δ = b2-4ac
Δ = 102-4·3·3
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-8}{2*3}=\frac{-18}{6} =-3 $
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+8}{2*3}=\frac{-2}{6} =-1/3 $

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