Hello, angela!
Could you be using your calculator incorrectly?
Yes, they plugged those numbers in, but I got a different answer than them.
We have:
.\(\displaystyle 2x^2\,-\,5x\,-\,4\:=\:0\qquad\Rightarrow\qquad a=2,\;b=-5,\;c=-4\)
Quadratic Formula:
. \(\displaystyle x\;=\;\frac{-(-5)\,\pm\,\sqrt{(-5)^2\,-\,4(2)(-4)}}{2(2)}\)
If you try to do all that on your calculator, you're bound to make errors.
I suggest you do a little simplifying first:
. \(\displaystyle x\;=\;\frac{5\,\pm\,\sqrt{57}}{4}\)
Even then, you can enter the problem incorrectly.
Let's calculate: \(\displaystyle \frac{5\, +\, \sqrt{57}}{4}\)
It looks like: \(\displaystyle [5]\;[+]\;[\sqrt{x}]\;[5][7]\;[\div]\;[4]\;[=]\) . . . but this is
wrong!
What we did was:
.\(\displaystyle 5\,+\,\frac{\sqrt{57}}{4}\)
. . (Order of operations: Division comes before Addition.)
We must divide the <u>entire</u> \(\displaystyle (5 + \sqrt{57})\) by \(\displaystyle 4.\)
We can add parentheses:
. \(\displaystyle [(]\:[5]\,[+]\,[\sqrt{x}]\,[5][7]\:[)]\,[\div]\,[4]\;[=]\)
. . . . . . . . . . . . . . . . . . . . . . .\(\displaystyle \Uparrow\)
. . . . . . . . . . . . . . . . . .\(\displaystyle \Uparrow\)
Or we can press \(\displaystyle [=]\) after the addition:
. . . \(\displaystyle [5]\,[+]\,[\sqrt{x}]\,[5][7]\,[=]\,[\div]\,[4]\;[=]\)
. . . . . . . . . . . . . . . . . . . . .\(\displaystyle \Uparrow\)