Answer:
8i and -8i
Step-by-step explanation:
Move everything into the left-hand side to get a quadratic equation:
[tex]\frac{1}{2}(x^2+40)=-12\\\frac{1}{2}x^2+20=-12\\\frac{1}{2}x^2+32=0[/tex]
Solve for x using the quadratic formula:
[tex]\frac{1}{2}x^2+32=0\\x=\frac{-b\±\sqrt{b^2-4ac}}{2a}\\x=\frac{-(0)\±\sqrt{0^2-4(\frac{1}{2})32}}{2(\frac{1}{2})}\\x=\±\sqrt{-64}\\x=\±8i[/tex]
The two imaginary solutions for the equation are 8i and -8i.
