Answer
$x_{1}=-\dfrac{1}{7}$ and $x_{2}=-1$
Work Step by Step
Given $7x^2+8x+1=0$
$a=7, \ b=8, \ c=1$
Using the quadratic formula: $\dfrac{-b\pm \sqrt{b^2-4ac}}{2a} , $ we have:
$\dfrac{-8\pm \sqrt{(8)^2-4\times 7\times 1}}{2\times 7} = \dfrac{-8\pm \sqrt{64-28}}{14} = \dfrac{-8\pm \sqrt{36}}{14} = \dfrac{-8\pm 6}{14}$
Therefore the solutions are $x_{1}=\dfrac{-8 + 6}{14}=\dfrac{-2}{14}=-\dfrac{1}{7}$ and $x_{2}=\dfrac{-8 - 6}{14}=\dfrac{-14}{14}=-1$