Solve for x: (2x-7)(x+4)=0?

Question

Solve for x: (2x-7)(x+4)=0

Answer 1

From (2x - 7)(x + 4) = 0, we get two possibilities,
(2x - 7) = 0 and (x + 4) = 0
x = 7/2 or x = -4

Answer 2

X=2
X=-1/4

Answer 3

x = 7 / 2 , x = - 4

Answer 4

(2x-7)(x+4)=0
for 2x - 7=0; we have 2x=7. Thus x =7/2 upon dividing both sides by 2.
for x+4=0; we have x =-4
Just manipulate the numbers. Remember its algebara. Whatever you do on one side also do it on the other. Like a scale, must always be balanced.

Answer 5

(2x - 7)(x + 4) = 0
(2x - 7) = 0 and (x + 4) = 0
x = 7/2, x = - 4

Answer: x = 3.5 or x = - 4

Proof:
Substitute x with 3.5:
(2[3.5] - 7) (3.5 + 4) = 0
(7 - 7) (3.5 + 4) = 0
0 * 7.5 = 0

Substitute x with 4:
(2[- 4] - 7) (- 4 + 4) = 0
(- 8 - 7) (0) = 0
- 15 * 0 = 0

Answer 6

2x-7=0 orx+4=0
x=7/2 x=-4

Answer 7

(2x-7)(x+4)=0 (multiply)
2x^2+8x-7x-28 (combine like terms)
2x^2+x-28=0 (this is a quadratic equation! u must provide with 2 solutions)

a=2, b=1, c=-28
formula: x=-b +or- square root of b^2 - 4(a)(c) over 2(a)

now substitute into formula:
a)-1 +or- square root of -1^2 - 4(2)(-28) over 2(2)
b)-1 +or- square root of 1 + 224 over 4
c)-1 +or- square root of 225 over 4
d)-1 +or- 15 over 4

solution#1:
-1 + 15 ---> 14
------------ = ---------- = 7/2
4 --------------> 4

solution#2:
-1 -15 --->-16
--------- = -------- = -4
4 ---------> 4

now substitute the 2 solutions [7/2 and -4]into the original equation:
2x^2+x-28=0
2(-4)^2 - 4 - 28=0
2(16) - 4 -28=0
32 -4 -28=0
28-28=0


solution #2 im not too good with fractions..lol

Answer 8

x=3.5,-4