5r^2-8r-5=0

Solution for 5r^2-8r-5=0 equation:

5r^2-8r-5=0

a = 5; b = -8; c = -5;
Δ = b2-4ac
Δ = -82-4·5·(-5)
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*5}=\frac{8-2\sqrt{41}}{10}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+2\sqrt{41}}{2*5}=\frac{8+2\sqrt{41}}{10}
$