0=9u^2-3u-16

Solution for 0=9u^2-3u-16 equation:

0=9u^2-3u-16

We move all terms to the left:

0-(9u^2-3u-16)=0

We add all the numbers together, and all the variables

-(9u^2-3u-16)=0

We get rid of parentheses

-9u^2+3u+16=0

a = -9; b = 3; c = +16;
Δ = b2-4ac
Δ = 32-4·(-9)·16
Δ = 585
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{585}=\sqrt{9*65}=\sqrt{9}*\sqrt{65}=3\sqrt{65}$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3\sqrt{65}}{2*-9}=\frac{-3-3\sqrt{65}}{-18}
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3\sqrt{65}}{2*-9}=\frac{-3+3\sqrt{65}}{-18}
$