5r2-10r-8=0

Solution for 5r2-10r-8=0 equation:

5r^2-10r-8=0

a = 5; b = -10; c = -8;
Δ = b2-4ac
Δ = -102-4·5·(-8)
Δ = 260
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:

$\sqrt{\Delta}=\sqrt{260}=\sqrt{4*65}=\sqrt{4}*\sqrt{65}=2\sqrt{65}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{65}}{2*5}=\frac{10-2\sqrt{65}}{10}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{65}}{2*5}=\frac{10+2\sqrt{65}}{10}
$