3q^2+10q+8=0

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Solution for 3q^2+10q+8=0 equation: 



3q^2+10q+8=0

a = 3; b = 10; c = +8;
Δ = b2-4ac
Δ = 102-4·3·8
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$


$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2}{2*3}=\frac{-12}{6}
=-2
$


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

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