3d^2-5d+2=0

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Solution for 3d^2-5d+2=0 equation: 



3d^2-5d+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:
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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


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


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

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