Determine the number and type of solutions for the equation \( 4 x^{2}-21 x+5=0 \) Two imaginary-number solutions One repeated irrational-number solution One repeated rational-number solution Two different irrational-number solutions Two different rational-number solutions Submit Question

To determine the number and type of solutions for the equation \( 4x^{2} - 21x + 5 = 0 \), we can utilize the discriminant from the quadratic formula, which is given by \( D = b^2 - 4ac \). Here, \( a = 4 \), \( b = -21 \), and \( c = 5 \). Calculating the discriminant gives us: \[ D = (-21)^2 - 4(4)(5) = 441 - 80 = 361 \] Since the discriminant \( D = 361 \) is positive and a perfect square, this indicates that there are two distinct rational-number solutions for the quadratic equation. In conclusion, the equation \( 4x^{2} - 21x + 5 = 0 \) has **two different rational-number solutions**.