r^2-2r-16=0

Solution for r^2-2r-16=0 equation:

r^2-2r-16=0

a = 1; b = -2; c = -16;
Δ = b2-4ac
Δ = -22-4·1·(-16)
Δ = 68
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:

$\sqrt{\Delta}=\sqrt{68}=\sqrt{4*17}=\sqrt{4}*\sqrt{17}=2\sqrt{17}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{17}}{2*1}=\frac{2-2\sqrt{17}}{2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{17}}{2*1}=\frac{2+2\sqrt{17}}{2}
$