what are the real solution to the equation [x]2+2[x]-3=0?

Question

if possible show me how u got ya answer

Answer 1

Assuming you mean x^2+2x-3 = 0

(x + 3)(x- 1) = 0

To get this, realize that the only way to get a -3 is to mulitiply 1 * 3 and one of them be negitive.

To figure out which one, you have to add them together and get 2 (the middle value) The only way to do that is 3 be positive and 1 be negitive.

Now, make each part equal to 0
x+3 = 0 x - 1 =0
x = -3 x = 1
So your solution is -3 or 1

Answer 2

2x + 2x - 3 = 0, you add the like terms, so now you have

4x - 3 =0 then to solve for x, add 3 to both sides so now you have

4x = 3 and then you divide by four, so then
x = 3/4 which is the same as .75

Answer 3

X^2+2X-3=0
(X+3)(X-1)=0

multiplying to have result=0 is 0

so first value
(X+3)=0
X=(-3)

second value
(X-1)=0
X=1


proof
X^2+2X-3=0
(-3)^2+2(-3)-3=0
9-6-3=0
9-9=0
0=0

X^2+2X-3=0
(1)^2+2(1)-3=0
1+2-3=0
3-3=0
0=0


so result x=1, x=-3

Answer 4

the equation factors nicely!
(x+3)(x-1) = 0
therefore
x = -3
or
x = 1

Answer 5

x= .75
.75 x 2 =1.50 + 2 x .75 = 1.50 - 3 =0

Answer 6

dan & ilove got it right

I got the same result using the quadratic formula