Me solve this problem, and someone clearly explain to me how to solve it.1.) use the value of the discriminant to determine if the given trinomials has 2 real solutions, 1 real solution, or no real solutions. a.  x2  − 4x  − 7 = 0b. 4r2  + 11r  − 3 = 0c. 3m2  + 7 = 0d.  t2  + 2t  + 1 = 0

Step-by-step explanation:

For a trinomial ax² + bx + c = 0, the discriminant is b² − 4ac.

If the discriminant is positive, there are 2 real solutions.

If the discriminant is 0, there is 1 real solution.

If the discriminant is negative, there are no real solutions.

a) x² − 4x − 7 = 0

Here, a = 1, b = -4, and c = -7.

b² − 4ac = (-4)² − 4(1)(-7) = 44

The discriminant is positive, so there are 2 real solutions.

b) 4r² + 11r − 3 = 0

Here, a = 4, b = 11, and c = -3.

b² − 4ac = (4)² − 4(11)(-3) = 148

The discriminant is positive, so there are 2 real solutions.

c) 3m² + 7 = 0

Here, a = 3, b = 0, and c = 7.

b² − 4ac = (0)² − 4(3)(7) = -84

The discriminant is negative, so there are no real solutions.

d) t² + 2t + 1 = 0

Here, a = 1, b = 2, and c = 1.

b² − 4ac = (2)² − 4(1)(1) = 0

The discriminant is zero, so there is 1 real solution.

the equivalent expression are always the hardest ones but you look at the gcf also

step-by-step explanation: