Answer
{${-\frac{3}{16}\pm\frac{3}{16}\sqrt (129)}$}
Work Step by Step
Step 1: Comparing $\frac{2}{3}x^{2}+\frac{1}{4}x−3=0$ to the standard form of a quadratic equation $ax^{2}+bx+c=0$;
$a=\frac{2}{3}$, $b=\frac{1}{4}$ and $c=−3$
Step 2: The quadratic formula is:
$x=\frac{-b\pm\sqrt (b^{2}-4ac)}{2a}$
Step 3: Substituting the values of a,b and c in the formula:
$x=\frac{-\frac{1}{4}\pm\sqrt ((\frac{1}{4})^{2}-4(\frac{2}{3})(-3))}{2(\frac{2}{3})}$
Step 4: $x=\frac{-\frac{1}{4}\pm\sqrt (\frac{1}{16}+8)}{\frac{4}{3}}$
Step 5: $x=\frac{-\frac{1}{4}\pm\sqrt (\frac{129}{16})}{\frac{4}{3}}$
Step 6: $x=\frac{-\frac{1}{4}\pm\sqrt (129\times\frac{1}{16})}{\frac{4}{3}}$
Step 7: $x=\frac{-\frac{1}{4}\pm(\sqrt (129)\sqrt \frac{1}{16})}{\frac{4}{3}}$
Step 8: $x=\frac{-\frac{1}{4}\pm\frac{1}{4}\sqrt (129)}{\frac{4}{3}}$
Step 9: $x=\frac{3}{4}({-\frac{1}{4}\pm\frac{1}{4}\sqrt (129)})$
Step 10: $x=\frac{3}{16}({-{1}\pm\sqrt (129)})$
Step 11: Therefore, the solution set is {${-\frac{3}{16}\pm\frac{3}{16}\sqrt (129)}$}.
