r^2-14r-169=0

Solution for r^2-14r-169=0 equation:

r^2-14r-169=0

a = 1; b = -14; c = -169;
Δ = b2-4ac
Δ = -142-4·1·(-169)
Δ = 872
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:

$\sqrt{\Delta}=\sqrt{872}=\sqrt{4*218}=\sqrt{4}*\sqrt{218}=2\sqrt{218}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-2\sqrt{218}}{2*1}=\frac{14-2\sqrt{218}}{2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+2\sqrt{218}}{2*1}=\frac{14+2\sqrt{218}}{2}
$