r^2+6=-5r

Solution for r^2+6=-5r equation:

r^2+6=-5r

We move all terms to the left:

r^2+6-(-5r)=0

We get rid of parentheses

r^2+5r+6=0

a = 1; b = 5; c = +6;
Δ = b2-4ac
Δ = 52-4·1·6
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-1}{2*1}=\frac{-6}{2}
=-3
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+1}{2*1}=\frac{-4}{2}
=-2
$