To determine the number of real solutions for the equation \( 4r^2 - 6r = 2 \), we first need to rewrite it in standard quadratic form, which is \( ax^2 + bx + c = 0 \).
Starting with the equation:
\[ 4r^2 - 6r - 2 = 0 \]
Here, we can identify the coefficients:
- \( a = 4 \)
- \( b = -6 \)
- \( c = -2 \)
Next, we calculate the discriminant \( D \) using the formula:
\[ D = b^2 - 4ac \]
Substituting in the values of \( a \), \( b \), and \( c \):
\[ D = (-6)^2 - 4(4)(-2) \] \[ D = 36 - 4 \times 4 \times -2 \] \[ D = 36 + 32 \] \[ D = 68 \]
Since the discriminant \( D \) is positive (\( 68 > 0 \)), this means that the quadratic equation has two distinct real solutions.
Thus, based on the value of the discriminant, the answer is:
Two real solutions.