Respuesta :

Answer:

[tex]\displaystyle x\approx -1.4\text{ or } 2.9[/tex]

Step-by-step explanation:

We want to solve the equation:

[tex]\displaystyle 2x^2-3x-8=0[/tex]

Using the quadratic formula.

The quadratic formula is given by:

[tex]\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]

In this case, a = 2, b = -3, and c = -8. Substitute:

[tex]\displaystyle x=\frac{-(-3)\pm\sqrt{(-3)^2-4(2)(-8)}}{2(2)}[/tex]

Evaluate:

[tex]\displaystyle x=\frac{3\pm\sqrt{73}}{4}[/tex]

So, our solutions are:

[tex]\displaystyle x=\frac{3+\sqrt{73}}{4}\approx 2.9\text{ and } x=\frac{3-\sqrt{73}}{4}\approx-1.4[/tex]

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