2r^2+8r+1=0

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

2r^2+8r+1=0

a = 2; b = 8; c = +1;
Δ = b2-4ac
Δ = 82-4·2·1
Δ = 56
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:

$\sqrt{\Delta}=\sqrt{56}=\sqrt{4*14}=\sqrt{4}*\sqrt{14}=2\sqrt{14}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-2\sqrt{14}}{2*2}=\frac{-8-2\sqrt{14}}{4}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+2\sqrt{14}}{2*2}=\frac{-8+2\sqrt{14}}{4}
$