#### Answer

$\color{blue}{\left\{2-\sqrt3, 2+\sqrt3\right\}}$.

#### Work Step by Step

Add $1$ to both sides of the equation to obtain:
$x^2-4x+1=0$
RECALL:
The quadratic equation $ax^2+bx+c=0$ can be solved using the quadratic formula:
$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$
The given equation has:
$a=1, b=-4, c=1$
Substitute these values into the quadratic formula to obtain:
$x=\dfrac{-(-4) \pm \sqrt{(-4)^2-4(1)(1)}}{2(1)}
\\x=\dfrac{4\pm\sqrt{16-4}}{2}
\\x=\dfrac{4\pm\sqrt{12}}{2}
\\x=\dfrac{4\pm\sqrt{4(3)}}{2}
\\x=\dfrac{4\pm2\sqrt{3}}{2}
\\x=2 \pm \sqrt3$
Thus, the solution set is $\color{blue}{\left\{2-\sqrt3, 2+\sqrt3\right\}}$.