r^2+15r+50=6-r

Solution for r^2+15r+50=6-r equation:

r^2+15r+50=6-r

We move all terms to the left:

r^2+15r+50-(6-r)=0

We add all the numbers together, and all the variables

r^2+15r-(-1r+6)+50=0

We get rid of parentheses

r^2+15r+1r-6+50=0

We add all the numbers together, and all the variables

r^2+16r+44=0

a = 1; b = 16; c = +44;
Δ = b2-4ac
Δ = 162-4·1·44
Δ = 80
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{80}=\sqrt{16*5}=\sqrt{16}*\sqrt{5}=4\sqrt{5}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-4\sqrt{5}}{2*1}=\frac{-16-4\sqrt{5}}{2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+4\sqrt{5}}{2*1}=\frac{-16+4\sqrt{5}}{2}
$