Find approximate solutions of the equation:
2x^2-7x-8=0

Question

anonymous · Answer

So,
a=2
b=-7
c=-8??

anonymous · Answer

Quadratic formula

anonymous · Answer

$$x = \frac{7 + \sqrt{113}}{4} , \frac{7 - \sqrt{113}}{4} $$

anonymous · Answer

So I end up with?
-(-7)$$\pm \sqrt{(-7)}^{2}-4*2*-8/2*2$$

anonymous · Answer

$$7\pm \sqrt{49}+64/4$$

anonymous · Answer

OK, OK...got it.  TY

anonymous · Answer

So before it said solve and wanted this form.  Now its asking for approxiamte solutions.  Is that something I can calculate out?

amistre64 · Answer

how close of an approx you need?

amistre64 · Answer

3 is an approx of pi

3.1 is better

3.1415 is even closer

3.14159  even better

all are approx.s

anonymous · Answer

Well it came up with -.91,4.41
I came up with 9.66,434
I must be punching it in the calculator wrong :(
But thank you for the definition of approximate

amistre64 · Answer

2x^2-7x-8=0

2x^2-7x = 8

x^2-7/2x = 4

x^2-(7/2)x + (7/4)^2 = 4 + (49/16)

(x-(7/4)^2 = 4 + (49/16)

x-(7/4) = sqrt(4 + (49/16))

x= (7/4) + sqrt(4 + (49/16)) =  4.407536
  = (7/4) - sqrt(4 + (49/16)) = -0.907536

id go with what it came up with :)