Help look at my photo

Step-by-step explanation:

One way to determine if the equation has any real solutions is to look at its discriminant. For the equation ax²+bx+c = 0, the discriminant is ...

  d = b² -4ac

When the discriminant is negative, both solutions to the quadratic are complex. There are no real solutions in that case.

We can find the discriminant values to be ...

  A: d = 2² -4(1)(4) = -12 . . . . no real zeros

  B: d = 0² -4(3)(-5) = 60 . . . two real zeros

  C: d = 8² -4(-2)(0) = 64 . . . two real zeros

  D: d = 10² -4(1)(26) = -4 . . .no real zeros

Expressions A and D have no real zeros.

Comment on the question

You are given an expression, not an equation. There is no equal sign. Hence, we cannot talk about solutions. We can only talk about zeros, values of x that make the expression have a value of zero.

this linear equation would have just one solution.

step-by-step explanation:

the reason why this would have one solution is that it is an expression with just one variable. since that variable is not raised to a power, it will always have just one solution.

it would take 5 hours i think idk i'm not that good at math.can you me with my questions i need asap.