Solve please: x^2+2x-2=0?

Question

also.. SOLVE

(5/x+6)-(2/x-1)=14/(x+6)(x-1)

Answer 1

PROBLEM 1:

a = 1
b = 2
c = -2

Quadratic formula:
..... -2 ± sqrt( 2² - 4(1)(-2) )
x = -------------------------------
................. 2

..... -2 ± sqrt( 4 + 8 )
x = -----------------------
................ 2

..... -2 ± sqrt( 12 )
x = -----------------------
................ 2

..... -2 ± 2 sqrt( 3 )
x = -----------------------
................ 2

x = -1 ± sqrt( 3 )

PROBLEM 2:

5 / (x+6) - 2 / (x-1) = 14 / (x + 6)(x - 1)

Multiply both sides by the common denominator of (x + 6)(x - 1):

5(x + 1) - 2(x + 6) = 14

Multiply out:
5x + 5 - 2x - 12 = 14

Group like terms:
3x - 7 = 14

Add 7 to both sides:
3x = 21
x = 21/3
x = 7

Answer 2

x^2 + 2x -- 2 = 0
=> x^2 + 2x + 1 = 3
=> (x + 1)^2 = 3
=> x + 1 = +/-- sqrt(3)
=> x = -- 1 + sqrt(3) or x = -- 1 -- sqrt(3)

5/(x + 6) -- 2/(x -- 1) = 14/(x + 6)(x -- 1)
multiplying both sides by (x + 6)(x -- 1),
5(x -- 1) -- 2(x + 6) = 14
=> 3x = 31
=> x = 31/3

Answer 3

ax^2 + bx + c = 0
is a quadratic which is solved with the quadratic formula:

x = [ -b +/- SQRT(b^2 - 4ac) ] / 2a

In your question, a = 1, b = 2 and c=2
b^2 - 4ac = 4 + 8 = 12

Since the determinant is greater than 0, there will be two solutions (that is why we have the "plus or minus" in front of the square root).

Answer 4

This isn't your Home Work is it? If it is, Shame on you!