(3x^2)+5x+2=0

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Solution for (3x^2)+5x+2=0 equation: 



(3x^2)+5x+2=0

a = 3; b = 5; c = +2;
Δ = b2-4ac
Δ = 52-4·3·2
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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


$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-1}{2*3}=\frac{-6}{6}
=-1
$


$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+1}{2*3}=\frac{-4}{6}
=-2/3
$

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