8(v)^(2)+18(v)+9=0

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Solution for 8(v)^(2)+18(v)+9=0 equation: 



8(v)^(2)+18(v)+9=0

a = 8; b = 18; c = +9;
Δ = b2-4ac
Δ = 182-4·8·9
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$


$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-6}{2*8}=\frac{-24}{16}
=-1+1/2
$


$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+6}{2*8}=\frac{-12}{16}
=-3/4
$

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