Respuesta :

xKelvin

Answer:

A

Step-by-step explanation:

For a quadratic equation in standard form:

[tex]ax^2+bx+c=0[/tex]

The discriminant, symoblized as Δ, is given by:

[tex]\Delta=b^2-4ac[/tex]

If:

  • Δ>0 (positive), then our quadratic has two real roots.
  • Δ<0 (negative), then our quadratic has no real roots. It does, however, have two imaginary (complex) roots.
  • And if Δ=0 (zero), then our quadratic has exactly one real root.

We have the equation:

[tex]-2x^2+x-28=0[/tex]

Thus:

[tex]a=-2, b=1, c=-28[/tex]

Substituting them for our discriminant:

[tex]\Delta=(1)^2-(4)(-2)(-28)[/tex]

Evaluate:

[tex]\Delta=1-224=-223<0[/tex]

Since our discriminant is negative, there is no real solutions.

Hence, our answer is A.

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