-15r^2+2r+1=0

Solution for -15r^2+2r+1=0 equation:

-15r^2+2r+1=0

a = -15; b = 2; c = +1;
Δ = b2-4ac
Δ = 22-4·(-15)·1
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-8}{2*-15}=\frac{-10}{-30}
=1/3
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+8}{2*-15}=\frac{6}{-30}
=-1/5
$