#### Answer

$\color{blue}{\left\{-\frac{5}{4}, 2\right\}}$

#### Work Step by Step

Divide $4$ to both sides:
$x^2-\frac{3}{4}x-\frac{10}{4}=0
\\x^2-\frac{3}{4}x -\frac{5}{2}=0$
Add $\frac{5}{2}$ to both sides:
$x-\frac{3}{4}x=\frac{5}{2}$
Add the square of one-half of the coefficient of $x$, which is $(\dfrac{-\frac{3}{4}}{2})^2=(\frac{-3}{8})^2=\frac{9}{64}$, to obtain:
$x^2-\frac{3}{4}x+\frac{9}{64}=\frac{5}{2}+\frac{9}{64}
\\(x-\frac{3}{8})^2=\frac{160}{64} + \frac{9}{64}
\\(x-\frac{3}{8})^2=\frac{169}{64}$
Take the square root of both sides to obtain:
$\sqrt{(x-\frac{3}{8})^2}=\pm \sqrt{\frac{169}{64}}
\\x-\frac{3}{8} = \pm \frac{13}{8}$
Add $\frac{3}{8}$ to both sides:
$x=\frac{3}{8} \pm \frac{13}{8}
\\x_1=\frac{3}{8}-\frac{13}{8}=\frac{-10}{8}=-\frac{5}{4}
\\x_2=\frac{3}{8}+\frac{13}{8}=\frac{16}{8}=2$
Thus, the solution set is: $\color{blue}{\left\{-\frac{5}{4}, 2\right\}}$.