0=r^2+8r-33

Solution for 0=r^2+8r-33 equation:

0=r^2+8r-33

We move all terms to the left:

0-(r^2+8r-33)=0

We add all the numbers together, and all the variables

-(r^2+8r-33)=0

We get rid of parentheses

-r^2-8r+33=0

We add all the numbers together, and all the variables

-1r^2-8r+33=0

a = -1; b = -8; c = +33;
Δ = b2-4ac
Δ = -82-4·(-1)·33
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-14}{2*-1}=\frac{-6}{-2}
=+3
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+14}{2*-1}=\frac{22}{-2}
=-11
$