8r^2-6-18r+1=0

Solution for 8r^2-6-18r+1=0 equation:

8r^2-6-18r+1=0

We add all the numbers together, and all the variables

8r^2-18r-5=0

a = 8; b = -18; c = -5;
Δ = b2-4ac
Δ = -182-4·8·(-5)
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-22}{2*8}=\frac{-4}{16}
=-1/4
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+22}{2*8}=\frac{40}{16}
=2+1/2
$