9j^2+33j+24=0

Solution for 9j^2+33j+24=0 equation:

9j^2+33j+24=0

a = 9; b = 33; c = +24;
Δ = b2-4ac
Δ = 332-4·9·24
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(33)-15}{2*9}=\frac{-48}{18}
=-2+2/3
$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(33)+15}{2*9}=\frac{-18}{18}
=-1
$