Answer
The solution set is $\left\{-\dfrac{-2-\sqrt2}{2}, \dfrac{2-\sqrt2}{2}\right\}$.
Work Step by Step
Subtract $-\frac{1}{2}$ on both sides of the equation to obtain:
$x^2+\sqrt{2}x-\frac{1}{2}=0$
RECALL:
A quadratic equation of the form $ax^2+bx+c=0$ can be solved using the quadratic formula
$x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$
The given quadratic equation has:
$a=1
\\b=\sqrt{2}
\\c=-\frac{1}{2}$
Substitute these values into the quadratic formula to obtain:
$z=\dfrac{-\sqrt2 \pm \sqrt{(\sqrt2)^2-4(1)(-\frac{1}{2})}}{2(1)}
\\z=\dfrac{-\sqrt2 \pm \sqrt{2-(-2)}}{2}
\\z=\dfrac{-\sqrt2\pm \sqrt{2+2}}{2}
\\x =\dfrac{-\sqrt2\pm\sqrt{4}}{2}
\\x=\dfrac{-\sqrt2\pm\sqrt{2^2}}{2}
\\x=\dfrac{-\sqrt2\pm 2}{2}$
Split the solutions to obtain:
$x_1 = \dfrac{-\sqrt2+2}{2}=\dfrac{2-\sqrt{2}}{2}
\\x_2=\dfrac{-\sqrt2-2}{2}=\dfrac{-2-\sqrt2}{2}$
Therefore, the solution set is $\left\{-\dfrac{-2-\sqrt2}{2}, \dfrac{2-\sqrt2}{2}\right\}$.
