Determine whether the values 10 and 30 are solutions to the quadratic equation (x−27)(x−10)=60.(1 point)

To determine whether the values \( x = 10 \) and \( x = 30 \) are solutions to the equation \( (x - 27)(x - 10) = 60 \), we can substitute each value into the equation.

First, let's check \( x = 10 \):

\[ (x - 27)(x - 10) = (10 - 27)(10 - 10) = (-17)(0) = 0 \]

Since \( 0 \neq 60 \), \( x = 10 \) is not a solution.

Next, let's check \( x = 30 \):

\[ (x - 27)(x - 10) = (30 - 27)(30 - 10) = (3)(20) = 60 \]

Since \( 60 = 60 \), \( x = 30 \) is a solution.

Conclusion:

• Only \( x = 30 \) is a solution.

So the correct response is: "Only \( x = 30 \) is a solution."