More Quadratic equations...?

Question

I solved the last one correctly, but I can't seem to get the correct answer on these others. I don't understand what I am doing wrong.

5x^2+4x-1=0

3x^2-4x=5

-x^2+3x+3=5

Here's what I did for problem #1:

5x^2+4x-1=0

a=5 b=4 c=-1

16+ sqrt 4^2-4(5)(-1)
over
2(5)

16+ sqrt 36
over
10

22
over
10

x=-2.2

Answer 1

5x² + 4x - 1 = 0

5x² - x + 5x - 1 = 0

x(5x - 1) + 1(5x - 1)

(x + 1)(5x - 1)

- - - - - - - - - - -s-

Answer 2

Problem #1 - Don't forget that there is a minus sign in front of the 1. So C in the quadratic formula is -1, not 1.

Problems 2 and 3 are not in standard form. Manipulate the problem so that everything is on one side and only zero on the other before applying the quadratic formula.

Problem 3 - The x^2 term has a minus sign, so A = -1. Be sure that your 4AC and 2A terms in the quadratic formula are correct.

Answer 3

5x^2+4x-1=0
s=4, p=-5
5x^2+ 5x - x -1 = 0
5x(x+1) - (x+1) = 0
(5x-1)(x+1) = 0

The best help is the example

Answer 4

using x = (-b ± sqrt(b^2 - 4ac))/(2a)

5x^2 + 4x - 1

(-4 ± sqrt(36))/10
(-4 ± 6)/10
(-10/10) or (2/10)
-1 or (1/5)

----------------------------------------

3x^2 - 4x = 5
3x^2 - 4x - 5 = 0

x = (4 ± sqrt(16 + 60))/6
x = (4 ± sqrt(76))/6
x = (4 ± 2sqrt(19))/6
x = (1/3)(2 ± sqrt(19))

-----------------------------------------

-x^2 + 3x + 3 = 5
-x^2 + 3x - 2 = 0
-(x^2 - 3x + 2) = 0
x^2 - 3x + 2 = 0
(x - 2)(x - 1) = 0
x = 2 or 1

Answer 5

1.5x^2+5x-x-1=0
5x(x+1)-1(x+1)=0
(5x-1)(x+1)=0
x=1/5 or -1

2.3x^2-4x-5=0
x=[4+/-(16+60)]/2*3
=[4+/-rt76]/6

3.x^2-3x+2=0
(x-2)(x-1)=0
x=1 or 2

Answer 6

ax² + bx + c = 0

-b +/- sqrt(b² - 4ac)
---------------------------
2a

1) -2/5 +/- sqrt(36)/10 = -2/5 +/- 3/5
x = 1/5 or -1

Can you take it from here?

Answer 7

what are you getting for solutions, and I will see if I can help?

Answer 8

why dont you post how you are doing it and i'll show you what you are doing wrong!