0=u^2-12u+9

Solution for 0=u^2-12u+9 equation:

0=u^2-12u+9

We move all terms to the left:

0-(u^2-12u+9)=0

We add all the numbers together, and all the variables

-(u^2-12u+9)=0

We get rid of parentheses

-u^2+12u-9=0

We add all the numbers together, and all the variables

-1u^2+12u-9=0

a = -1; b = 12; c = -9;
Δ = b2-4ac
Δ = 122-4·(-1)·(-9)
Δ = 108
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:

$\sqrt{\Delta}=\sqrt{108}=\sqrt{36*3}=\sqrt{36}*\sqrt{3}=6\sqrt{3}$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-6\sqrt{3}}{2*-1}=\frac{-12-6\sqrt{3}}{-2}
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+6\sqrt{3}}{2*-1}=\frac{-12+6\sqrt{3}}{-2}
$