Question

I'm stuck on this last question and i'm not sure what i'm doing wrong, I can't seem to find the right solution.

2x^2 - 3x - 4 = 0

From the Quadratic Formula
-b ± √(b² - 4ac))/2a
-3 ± √(-3^2 - 4(2)(4))/2(2)
-3 ± √(-9 - 32)/4
(everything else above this point I think I did correctly, but I got stuck here, can someone help me?)
-3 ± √(41)/4
-3 ± 6.4/4

This is what the solution should be: x ≈ 2.35 or x ≈ -0.85

Answer 1

\[

x=\frac{1}{4}
\left(3-\sqrt{41}\right)\\

y=\frac{1}{4}
\left(3+\sqrt{41}\right)

\]

Answer 2

I'm afraid I still don't understand??

Answer 3

@Renee99
b=-3
b^2 = 9 and not -9
This might be where you sent wrong.

Answer 4

ohhh

Answer 5

Also
-4( a c) =(-4) (2)(-4)= 32 and -32

Answer 6

ohh, okay I see now, let me redo it and see if I get it right this time :)

Answer 7

also
-b =3 and not -3

Answer 8

Use quadratic formula

x = [-b + √(b² - 4ac)] / 2a
or
x = [-b - √(b² - 4ac)] / 2a

So using the values
a = 2
b = -3
c = -4

x = [-(-3) + √((-3)² - (4)(2)(-4)] / 2(2)
or
x = [-(-3) - √((-3)² - (4)(2)(-4)] / 2(2)

so

x = -[√(41)-3] / 4,
or
x = [√(41)+3] / 4

Answer 9

Best answer me(:

Answer 10

@Math4Life you should not ask for Best answer me. One should leav it to the asker.

Answer 11

This is what I have now:
-b ± √(b^2 - 4ac))/2a
3 ± √(-3^2 - 4(2)(-4))/2(2)
3 ± √(9 + 32)/4
3 ± √(41)/4
3 ± 10.25/4

Answer 12

Yes, this is right.

Answer 13

okay, so now how do I get to this: x ≈ 2.35 or x ≈ -0.85

Answer 14

Use your calculator to compute
\[x=\frac{1}{4}
\left(3-\sqrt{41}\right)=-0.85\\

y=\frac{1}{4}
\left(3+\sqrt{41}\right)=2.35
\]

Answer 15

@eliassaab ok sorry

Answer 16

but this is your whole problem if you want it @Renee99

Use quadratic formula

x = [-b + √(b² - 4ac)] / 2a
or
x = [-b - √(b² - 4ac)] / 2a

So using the values
a = 2
b = -3
c = -4

x = [-(-3) + √((-3)² - (4)(2)(-4)] / 2(2)
or
x = [-(-3) - √((-3)² - (4)(2)(-4)] / 2(2)

so

x = -[√(41)-3] / 4,
or
x = [√(41)+3] / 4

Answer 17

okay, phew.... gotta take a breather after that one, lol! Anyway, thanks for both yours help! I really appreciate it! :)