Answer
The equation has two real solutions.
Work Step by Step
$\frac{1}{2}x^{2}=7x-1\implies\frac{1}{2}x^{2}-7x+1=0$
Comparing $\frac{1}{2}x^{2}-7x+1=0$ with $ax^{2}+bx+c$, we see that
$a=\frac{1}{2}, b=-7$ and $c=1$.
$b^{2}-4ac=(-7)^{2}-4(\frac{1}{2})(1)=47$.
The discriminant is greater than $0$. So, the equation has two real solutions.
