• Nov 28th 2008, 09:06 AM
Joker37
Use the quadratic formula to solve:

-4x^2 + 13x - 3 = 0
• Nov 28th 2008, 09:09 AM
brentwoodbc
# infront of x^2 is a
infront on x is b
infront of nothing constant term is c

also make sure equation is in standard form which it is.

then fill into

x= -b sqrt (b^2 -4ac)
2a
• Nov 28th 2008, 09:13 AM
Joker37
Quote:

Originally Posted by brentwoodbc
# infront of x^2 is a
infront on x is b
infront of nothing constant term is c

also make sure equation is in standard form which it is.

then fill into

x= -b sqrt (b^2 -4ac)
2a

Yes, I tried that but I didn't get the right answer.
• Nov 28th 2008, 09:14 AM
brentwoodbc
make sure you keep track of negatives "2 neg. = pos."
also what is the given answer?
• Nov 28th 2008, 09:17 AM
brentwoodbc
also its + the sqrt sorry so -b plus or minus the sqrt
• Nov 28th 2008, 09:20 AM
brentwoodbc
I get x=1/4,3
• Nov 28th 2008, 09:23 AM
Joker37
I got 2 ± √6 instead of 7 ± 5√2.

EDIT: Sorry, I meant I got -13 ± √217/ -8 instead of 9 ± √89/ 2.
• Nov 28th 2008, 09:28 AM
brentwoodbc
Quote:

Originally Posted by Joker37
I got 2 ± √6 instead of 7 ± 5√2.

thats odd maybe someone else can chime in on this but I get

-13 +/- √121
-8

and 121 is a perfect square so

-13 +/- 11
-8

so

1/4 , 3