0=u^2-19u-216

Solution for 0=u^2-19u-216 equation:

0=u^2-19u-216

We move all terms to the left:

0-(u^2-19u-216)=0

We add all the numbers together, and all the variables

-(u^2-19u-216)=0

We get rid of parentheses

-u^2+19u+216=0

We add all the numbers together, and all the variables

-1u^2+19u+216=0

a = -1; b = 19; c = +216;
Δ = b2-4ac
Δ = 192-4·(-1)·216
Δ = 1225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1225}=35$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-35}{2*-1}=\frac{-54}{-2}
=+27
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+35}{2*-1}=\frac{16}{-2}
=-8
$