r^2-8r-25=0

Solution for r^2-8r-25=0 equation:

r^2-8r-25=0

a = 1; b = -8; c = -25;
Δ = b2-4ac
Δ = -82-4·1·(-25)
Δ = 164
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:

$\sqrt{\Delta}=\sqrt{164}=\sqrt{4*41}=\sqrt{4}*\sqrt{41}=2\sqrt{41}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-2\sqrt{41}}{2*1}=\frac{8-2\sqrt{41}}{2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+2\sqrt{41}}{2*1}=\frac{8+2\sqrt{41}}{2}
$