3r^2+11r+6=0

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Solution for 3r^2+11r+6=0 equation: 



3r^2+11r+6=0

a = 3; b = 11; c = +6;
Δ = b2-4ac
Δ = 112-4·3·6
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$


$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-7}{2*3}=\frac{-18}{6}
=-3
$


$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+7}{2*3}=\frac{-4}{6}
=-2/3
$

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