9j^2+18j+8=0

Solution for 9j^2+18j+8=0 equation:

9j^2+18j+8=0

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

$\sqrt{\Delta}=\sqrt{36}=6$
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-6}{2*9}=\frac{-24}{18}
=-1+1/3
$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+6}{2*9}=\frac{-12}{18}
=-2/3
$