3r^2+10r+3=0

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Solution for 3r^2+10r+3=0 equation: 



3r^2+10r+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:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$


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


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

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