Which equation has the solutions x = 1+
5?
x2 + 2x + 4 = 0
x2 - 2x + 4 = 0
0x²+2x–4=0
x2 - 2x -4 = 0

The last equation x2 - 2x -4 = 0

has solution  (x - 1)^2 - 5 = 0,   x  =  1 + root(5)  or  x = 1 - root(5)

Step-by-step explanation:

If a quadratic function has roots 1 and 5

f(x) = (x -1)(x- 5)

f(x) = x^2 - 6x + 5

Unless you meant.  -4 and 6  ?

g(x) = (x + 4)(x - 6)

g(x) = x^2 -2x -24

Or did you mean  x = 1 and x =4 ?...

x^2 + 2x + 4 = 0  :   complete square  x^2 + 2x + 1 + 3 = 0,   (x+1)^2 + 3 = 0

x^2 - 2x + 4 = 0 :  complete square:  (x -1)^2 + 3 = 0

0x^2 + 2x - 4 = 0,    2x - 4 = 0,  x = 2

x^2 - 2x - 4 = 0  becomes:   x^2 - 2x + 1 - 1 -4 = 0 ;  (x - 1)^2 - 5 = 0

step-by-step explanation:

ok one would be money, and the other would be refreshments or something similar. if each bag of popcorn is 5.00 and each drink is half of that that means drinks are 2.50 the proper way to do that would to look and see how much of each you can buy without exceeding a limit of 30.00

"standard form:

3x2 + 14x − 10 = 0

solutions based on quadratic formula:

x1   = −14 − √ 142 − 4×3×(−10) /   2×3   = −14 − 2 × √ 79 /   6   ≈ −5.296

x2   = −14 + √ 142 − 4×3×(−10) /   2×3   = −14 + 2 × √ 79 /   6   ≈ 0.629