r^2+6r-16=50

Solution for r^2+6r-16=50 equation:

r^2+6r-16=50

We move all terms to the left:

r^2+6r-16-(50)=0

We add all the numbers together, and all the variables

r^2+6r-66=0

a = 1; b = 6; c = -66;
Δ = b2-4ac
Δ = 62-4·1·(-66)
Δ = 300
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:

$\sqrt{\Delta}=\sqrt{300}=\sqrt{100*3}=\sqrt{100}*\sqrt{3}=10\sqrt{3}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-10\sqrt{3}}{2*1}=\frac{-6-10\sqrt{3}}{2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+10\sqrt{3}}{2*1}=\frac{-6+10\sqrt{3}}{2}
$