8y^2+y=0

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Solution for 8y^2+y=0 equation: 



8y^2+y=0

a = 8; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·8·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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


$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-1}{2*8}=\frac{-2}{16}
=-1/8
$


$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+1}{2*8}=\frac{0}{16}
=0
$

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