text: # How do you solve 45-sqrt(10-2x^2)=25?
  
  Jun 29, 2017
  
  #### Answer:
  
  See a solution process below:
  
  #### Explanation:
  
  First, subtract $\textcolor{red}{45}$ from each side of the equation to isolate the radical term while keeping the equation balanced:
  
  $- \textcolor{red}{45} + 45 - \sqrt{10 - 2 {x}^{2}} = - \textcolor{red}{45} + 25$
  
  $0 - \sqrt{10 - 2 {x}^{2}} = - 20$
  
  $- \sqrt{10 - 2 {x}^{2}} = - 20$
  
  Next, square both sides of the equation to eliminate the radical while keeping the equation balanced:
  
  ${\left(- \sqrt{10 - 2 {x}^{2}}\right)}^{2} = {\left(- 20\right)}^{2}$
  
  $10 - 2 {x}^{2} = 400$
  
  Then, subtract $\textcolor{red}{10}$ from each side of the equation to isolate the $x$ term while keeping the equation balanced:
  
  $- \textcolor{red}{10} + 10 - 2 {x}^{2} = - \textcolor{red}{10} + 400$
  
  $0 - 2 {x}^{2} = 390$
  
  $- 2 {x}^{2} = 390$
  
  Then, divide each side of the equation by $\textcolor{red}{- 2}$ to isolate ${x}^{2}$ while keeping the equation balanced:
  
  $\left(- 2 {x}^{2}\right) \textcolor{red}{- 2} = \frac{390}{\textcolor{red}{- 2}}$
  
  $\left(\textcolor{red}{\cancel{\textcolor{b l a c k}{- 2}}} {x}^{2}\right) \cancel{\textcolor{red}{- 2}} = - 195$
  
  ${x}^{2} = - 195$
  
  Because any number squared always produces a positive result, there is no solution for $x$ which will result in a negative 195.
  
  Or, the solution is the null or empty set: $\left\{\emptyset\right\}$