4r^2-33=-r

Solution for 4r^2-33=-r equation:

4r^2-33=-r

We move all terms to the left:

4r^2-33-(-r)=0

We add all the numbers together, and all the variables

4r^2-(-1r)-33=0

We get rid of parentheses

4r^2+1r-33=0

We add all the numbers together, and all the variables

4r^2+r-33=0

a = 4; b = 1; c = -33;
Δ = b2-4ac
Δ = 12-4·4·(-33)
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{529}=23$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-23}{2*4}=\frac{-24}{8}
=-3
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+23}{2*4}=\frac{22}{8}
=2+3/4
$