Answer
$\frac{4}{13}\approx 0.308, \ or, \ -14$
Work Step by Step
For the equation $13x^2 +178x-56 = 0$, we use the quadratic formula as follows
$$
x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a},
\\
a=13, \quad b=178, \quad c=-56
.$$
That is, $$ x=\frac{-178 \pm \sqrt{178^2-4(13)(-56)}}{2(13) }=\frac{4}{13}\approx 0.308, \ or, \ -14
.$$
Check the result:
$$
13(0.308)^2 +178(0.308)-56 \approx 56-56=0, \quad
13(-14)^2 +178(-14)-56 =56-56=0
.
$$
