r^2-10r-5=0

Solution for r^2-10r-5=0 equation:

r^2-10r-5=0

a = 1; b = -10; c = -5;
Δ = b2-4ac
Δ = -102-4·1·(-5)
Δ = 120
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:

$\sqrt{\Delta}=\sqrt{120}=\sqrt{4*30}=\sqrt{4}*\sqrt{30}=2\sqrt{30}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{30}}{2*1}=\frac{10-2\sqrt{30}}{2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{30}}{2*1}=\frac{10+2\sqrt{30}}{2}
$