3x2+8x+1=0.

Solution for 3x2+8x+1=0. equation:

3x^2+8x+1=0.

a = 3; b = 8; c = +1;
Δ = b2-4ac
Δ = 82-4·3·1
Δ = 52
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{52}=\sqrt{4*13}=\sqrt{4}*\sqrt{13}=2\sqrt{13}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-2\sqrt{13}}{2*3}=\frac{-8-2\sqrt{13}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+2\sqrt{13}}{2*3}=\frac{-8+2\sqrt{13}}{6}
$