8d2+19d+6=0

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Solution for 8d2+19d+6=0 equation:



8d^2+19d+6=0
a = 8; b = 19; c = +6;
Δ = b2-4ac
Δ = 192-4·8·6
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-13}{2*8}=\frac{-32}{16} =-2 $
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+13}{2*8}=\frac{-6}{16} =-3/8 $

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