-u^2=+12u-29=0

Solution for -u^2=+12u-29=0 equation:

-u^2=+12u-29=0

We move all terms to the left:

-u^2-(+12u-29)=0

We add all the numbers together, and all the variables

-u^2-(12u-29)=0

We add all the numbers together, and all the variables

-1u^2-(12u-29)=0

We get rid of parentheses

-1u^2-12u+29=0

a = -1; b = -12; c = +29;
Δ = b2-4ac
Δ = -122-4·(-1)·29
Δ = 260
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:

$\sqrt{\Delta}=\sqrt{260}=\sqrt{4*65}=\sqrt{4}*\sqrt{65}=2\sqrt{65}$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-2\sqrt{65}}{2*-1}=\frac{12-2\sqrt{65}}{-2}
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+2\sqrt{65}}{2*-1}=\frac{12+2\sqrt{65}}{-2}
$