3r+r^2=0

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



3r+r^2=0

a = 1; b = 3; c = 0;
Δ = b2-4ac
Δ = 32-4·1·0
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$


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


$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3}{2*1}=\frac{0}{2}
=0
$

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