How do you solve -x^2-x+1=0 in a quadratic formula? !

Step-by-step explanation:

For a quadratic equation ax² + bx + c = 0, x can be found with the quadratic formula:

x = [ -b ± √(b² − 4ac) ] / 2a

-x² − x + 1 = 0

Here, a = -1, b = -1, and c = 1.

x = [ -(-1) ± √((-1)² − 4(-1)(1)) ] / 2(-1)

x = [ 1 ± √(1 + 4) ] / -2

x = (1 ± √5) / -2

x = (-1 ± √5) / 2

The two values of x are (-1 − √5) / 2 and (-1 + √5) / 2.

answer: .64

step-by-step explanation

set up the formula for standard deviation for this set of numbers, then simplify

{x = -1/2 ,   y = -1

step-by-step explanation:

solve the following system:

{2 x + y = -2

{10 x - 3 y = -2

in the first equation, look to solve for y:

{2 x + y = -2

{10 x - 3 y = -2

subtract 2 x from both sides:

{y = -2 x - 2

{10 x - 3 y = -2

substitute y = -2 x - 2 into the second equation:

{y = -2 x - 2

{10 x - 3 (-2 x - 2) = -2

10 x - 3 (-2 x - 2) = (6 x + 6) + 10 x = 16 x + 6:

{y = -2 x - 2

{16 x + 6 = -2

in the second equation, look to solve for x:

{y = -2 x - 2

{16 x + 6 = -2

subtract 6 from both sides:

{y = -2 x - 2

{16 x = -8

divide both sides by 16:

{y = -2 x - 2

{x = -1/2

substitute x = -1/2 into the first equation:

{y = -1

{x = -1/2

collect results in alphabetical order:

answer:   {x = -1/2 ,   y = -1