40+13r=-r^2

Solution for 40+13r=-r^2 equation:

40+13r=-r^2

We move all terms to the left:

40+13r-(-r^2)=0

We get rid of parentheses

r^2+13r+40=0

a = 1; b = 13; c = +40;
Δ = b2-4ac
Δ = 132-4·1·40
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-3}{2*1}=\frac{-16}{2}
=-8
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+3}{2*1}=\frac{-10}{2}
=-5
$