3d^2+93d=0

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Solution for 3d^2+93d=0 equation: 



3d^2+93d=0

a = 3; b = 93; c = 0;
Δ = b2-4ac
Δ = 932-4·3·0
Δ = 8649
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{8649}=93$


$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(93)-93}{2*3}=\frac{-186}{6}
=-31
$


$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(93)+93}{2*3}=\frac{0}{6}
=0
$

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