Question

Find the number of real number solutions for the equation. x^2-3x+8=0

Answer 1

`x^2-3x+8=0 \\\\\Delta=(-3)^2-4\cdot1\cdot8=9-32=-23`

`\Delta<0` therefore, there are no real solutions.

Answer 2

None.

Explanation:

The coefficients are a=1, b=-3, c=8, so the discriminant is ...

... b²-4ac = (-3)²-4(1)(8) = 9-32 = -23

This negative value means both the roots are complex.