3r^2-8r-5=0

Solution for 3r^2-8r-5=0 equation:

3r^2-8r-5=0

a = 3; b = -8; c = -5;
Δ = b2-4ac
Δ = -82-4·3·(-5)
Δ = 124
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:

$\sqrt{\Delta}=\sqrt{124}=\sqrt{4*31}=\sqrt{4}*\sqrt{31}=2\sqrt{31}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-2\sqrt{31}}{2*3}=\frac{8-2\sqrt{31}}{6}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+2\sqrt{31}}{2*3}=\frac{8+2\sqrt{31}}{6}
$