Answer
The equation has two real solutions.
Work Step by Step
Given: $-\frac{1}{4}x^{2}+4x=-2$
$\implies -\frac{1}{4}x^{2}+4x+2=0$
Comparing $-\frac{1}{4}x^{2}+4x+2=0$ with $ax^{2}+bx+c$, we see that
$a=-\frac{1}{4}, b=4$ and $c=2$.
$b^{2}-4ac=(4)^{2}-4(-\frac{1}{4})(2)=18$.
The discriminant is greater than $0$. So, the equation has two real solutions.
