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]
